1. Problem Solving Tutor Next This presentation is designed to develop your problem solving skills in quantitative chemistry. Working through the whole tutor program you will develop and use three techniques for problem solving. These have been called: 1. Knowing what you know (do this one first) 2. Tabulating the information (do this one second) 3. Storyboarding (do this one last) You are about to start: Storyboarding

2. Some important points: You will need to use a calculator (or the calculator function on your computer/device). Do not use the keyboard to advance or go back through slides. Use the mouse/screen tap and the icons: NextBack Next

3. Problem solving tutor Storyboarding – a method of making complex experimental descriptions clear. Next

4. Storyboarding Storyboarding is a good technique to use on questions with a detailed experimental procedure. You sketch out a drawing of the method then label it with relevant quantities and the calculation. Here we will model on screen what we want you to be able to do on paper. Next

5. Question 1 Storyboarding 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.100 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Next

6. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. A Click on the piece of apparatus shown on the right that should go in the space marked by A, then click Next. Next

7. Apparatus feedback A You make up a solution to a certain volume in a volumetric flask. Back

8. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. A B Click on the piece of apparatus shown on the right that should go in the space marked by A, then click Next. Next

9. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. B Click on the piece of apparatus shown on the right that should go in the space marked by B, then click Next. B Next

10. Apparatus feedback B To take out an exact volume like 10 or 25 cm 3 of the solution we would use a pipette. Back

11. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. B Click on the piece of apparatus shown on the right that should go in the space marked by B, then click Next. Next B

12. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. C Click on the piece of apparatus shown on the right that should go in the space marked by C, then click Next. Next

13. Apparatus feedback C We use a conical flask for titrations because the narrow neck makes it easy to swirl without spilling any of the contents. Back

14. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. C Click on the piece of apparatus shown on the right that should go in the space marked by C, then click Next. Next

15. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the label shown on the right that would be best to go in the space marked by 1, then click Next. Next Made up to 250 cm 3 of diluted HCl(aq) 0.100 mol dm -3 NaOH(aq) 25 cm 3 removed is 1/10 of the dilute solution 19.8 cm 3 of NaOH added to reach end point 10 cm 3 of original HCl(aq) 1 2 3 4 5 NaOH was added to reach end point

16. Label feedback 1 Here we are taking the 10 cm 3 of the original acid. Notice that the labels should contain the quantities if appropriate. Back

17. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the label shown on the right that would be best to go in the space marked by 1, then click Next. Next Made up to 250 cm 3 of diluted HCl(aq) 0.100 mol dm -3 NaOH(aq) 25 cm 3 removed is 1/10 of the dilute solution 19.8 cm 3 of NaOH added to reach end point 10 cm 3 of original HCl(aq) 1 2 3 4 5 NaOH was added to reach end point

18. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the label shown on the right that would be best to go in the space marked by 2, then click Next. Made up to 250 cm 3 of diluted HCl(aq) 0.100 mol dm -3 NaOH(aq) 25 cm 3 removed is 1/10 of the dilute solution 19.8 cm 3 of NaOH added to reach end point 10 cm 3 of original HCl(aq) 2 3 4 5 NaOH was added to reach end point Next

19. Label feedback 2 Here we are diluting the 10 cm 3 of the original acid up to 250 cm 3. Notice that the labels should contain the quantities if appropriate. Back

20. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the label shown on the right that would be best to go in the space marked by 2, then click Next. Next Made up to 250 cm 3 of diluted HCl(aq) 0.100 mol dm -3 NaOH(aq) 25 cm 3 removed is 1/10 of the dilute solution 19.8 cm 3 of NaOH added to reach end point 2 3 4 5 NaOH was added to reach end point 10 cm 3 of original HCl(aq)

21. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the label shown on the right that would be best to go in the space marked by 3, then click Next. Next 0.100 mol dm -3 NaOH(aq) 25 cm 3 removed is 1/10 of the dilute solution 19.8 cm 3 of NaOH added to reach end point 2 3 4 5 NaOH was added to reach end point Made up to 250 cm 3 of diluted HCl(aq) 10 cm 3 of original HCl(aq)

22. Label feedback 3 Here we are taking 25 cm 3 of the 250 cm 3 of dilute acid. Notice that the labels should contain the quantities, such as: this is one-tenth of the whole. Back

23. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the label shown on the right that would be best to go in the space marked by 3, then click Next. Next 0.100 mol dm -3 NaOH(aq) 25 cm 3 removed is 1/10 of the dilute solution 19.8 cm 3 of NaOH added to reach end point 2 3 4 5 NaOH was added to reach end point Made up to 250 cm 3 of diluted HCl(aq) 10 cm 3 of original HCl(aq)

24. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the label shown on the right that would be best to go in the space marked by 4, then click Next. 0.100 mol dm -3 NaOH(aq) 19.8 cm 3 of NaOH added to reach end point 2 3 4 5 NaOH was added to reach end point Made up to 250 cm 3 of diluted HCl(aq) 10 cm 3 of original HCl(aq) 25 cm 3 removed is 1/10 of the dilute solution Next

25. Label feedback 4 Here we are labelling the contents of the burette. Notice that the labels should contain the quantities such as the concentration of the NaOH. Back

26. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the label shown on the right that would be best to go in the space marked by 4, then click Next. Next 0.100 mol dm -3 NaOH(aq) 19.8 cm 3 of NaOH added to reach end point 2 3 4 5 NaOH was added to reach end point Made up to 250 cm 3 of diluted HCl(aq) 10 cm 3 of original HCl(aq) 25 cm 3 removed is 1/10 of the dilute solution

27. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the label shown on the right that would be best to go in the space marked by 5, then click Next. 19.8 cm 3 of NaOH added to reach end point 2 3 4 5 NaOH was added to reach end point Made up to 250 cm 3 of diluted HCl(aq) 10 cm 3 of original HCl(aq) 25 cm 3 removed is 1/10 of the dilute solution 0.100 mol dm -3 NaOH(aq) Next

28. Label feedback 5 It is better to state the titre value in the label – the information you are going to use in the calculation should all appear on the storyboard. Back

29. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the label shown on the right that would be best to go in the space marked by 5, then click Next. Next 19.8 cm 3 of NaOH added to reach end point 2 3 4 5 NaOH was added to reach end point Made up to 250 cm 3 of diluted HCl(aq) 10 cm 3 of original HCl(aq) 25 cm 3 removed is 1/10 of the dilute solution 0.100 mol dm -3 NaOH(aq)

30. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the correct quantity for the number of moles of NaOH, then click Next. Next 2 3 4 Made up to 250 cm 3 of diluted HCl(aq) 10 cm 3 of original HCl(aq) 25 cm 3 removed is 1/10 of the dilute solution 0.100 mol dm -3 NaOH(aq) 19.8 1.98 0.198 0.0198 0.00198 0.000198 19.8 cm 3 of NaOH added to reach end point = mol

31. Moles of NaOH feedback To calculate the number of moles of NaOH use moles = concentration x volume (in dm 3 ). Back

32. 19.8 cm 3 of NaOH added to reach end point = mol Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the correct quantity for the number of moles of NaOH, then click Next. Next 2 3 4 Made up to 250 cm 3 of diluted HCl(aq) 10 cm 3 of original HCl(aq) 25 cm 3 removed is 1/10 of the dilute solution 0.100 mol dm -3 NaOH(aq) 19.8 1.98 0.198 0.0198 0.00198 0.000198 19.8 cm 3 of NaOH added to reach end point = 0.00198 mol of NaOH

33. Moles of HCl = Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the correct quantity for the number of moles of HCl in the conical flask, then click Next. Next 2 3 4 Made up to 250 cm 3 of diluted HCl(aq) 10 cm 3 of original HCl(aq) 25 cm 3 removed is 1/10 of the dilute solution 0.100 mol dm -3 NaOH(aq) 19.8 1.98 0.198 0.0198 0.00198 0.000198 19.8 cm 3 of NaOH added to reach end point = 0.00198 mol of NaOH

34. Moles of HCl feedback 1 To calculate the number of moles of HCl use the mole ratio and the number of moles of NaOH. The equation is: HCl + NaOH  NaCl + H 2 O Back

35. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the correct quantity for the number of moles of HCl in the conical flask, then click Next. Next 2 3 4 Made up to 250 cm 3 of diluted HCl(aq) 10 cm 3 of original HCl(aq) 25 cm 3 removed is 1/10 of the dilute solution 0.100 mol dm -3 NaOH(aq) 19.8 1.98 0.198 0.0198 0.000198 Mol of HCl = 19.8 cm 3 of NaOH added to reach end point = 0.00198 mol of NaOH 0.00198 Moles of HCl = 0.00198 (in 25 cm 3 of dilute solution)

36. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the correct quantity for the number of moles of HCl in the volumetric flask, then click Next. Next 3 4 10 cm 3 of original HCl(aq) 25 cm 3 removed is 1/10 of the dilute solution 0.100 mol dm -3 NaOH(aq) 19.8 1.98 0.198 0.0198 0.000198 19.8 cm 3 of NaOH added to reach end point = 0.00198 mol of NaOH 0.00198 Moles of HCl = 0.00198 (in 25 cm 3 of dilute solution) Made up to 250 cm 3 of diluted HCl(aq) = 0.0198 mol Made up to 250 cm 3 of diluted HCl(aq) = mol

37. Moles of HCl feedback 2 25 cm 3, which is 1/10 of the 250 cm 3 solution, was taken and titrated against NaOH. So to work backwards from the amount of HCl in the conical flask you need to multiply by 10. Back

38. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the correct quantity for the number of moles of HCl in the volumetric flask, then click Next. Next 3 4 10 cm 3 of original HCl(aq) 25 cm 3 removed is 1/10 of the dilute solution 0.100 mol dm -3 NaOH(aq) 19.8 1.98 0.198 0.0198 0.000198 19.8 cm 3 of NaOH added to reach end point = 0.00198 mol of NaOH 0.00198 Moles of HCl = 0.00198 (in 25 cm 3 of dilute solution) Made up to 250 cm 3 of diluted HCl(aq) = 0.0198 mol

39. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the correct quantity for the original concentration of HCl, then click Next. Next 2 3 4 25 cm 3 removed is 1/10 of the dilute solution 0.100 mol dm -3 NaOH(aq) 19.8 1.98 0.198 0.0198 0.000198 19.8 cm 3 of NaOH added to reach end point = 0.00198 mol of NaOH 0.00198 Moles of HCl = 0.00198 (in 25 cm 3 of dilute solution) 10 cm 3 of original HCl(aq) conc = M Made up to 250 cm 3 of diluted HCl(aq) = 0.0198 mol

40. Concentration of HCl feedback The number of moles of HCl in the 250 cm 3 of dilute solution = the number of moles of HCl in 10 cm 3 of the original acid. Use concentration = moles / volume for this 10 cm 3 of the original acid. Back

41. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the correct quantity for the original concentration of HCl, then click Next. Next 2 3 4 25 cm 3 removed is 1/10 of the dilute solution 0.100 mol dm -3 NaOH(aq) 0.198 0.0198 0.000198 19.8 cm 3 of NaOH added to reach end point = 0.00198 mol of NaOH 0.00198 Moles of HCl = 0.00198 (in 25 cm 3 of dilute solution) Made up to 250 cm 3 of diluted HCl(aq) = 0.0198 mol 10 cm 3 of original HCl(aq) conc = M 19.8 1.98

42. Q1. 10 cm 3 of hydrochloric acid of unknown concentration was put into a 250 cm 3 volumetric flask and it was made up to the mark with distilled water. 25.0 cm 3 of this diluted hydrochloric acid required 19.8 cm 3 of 0.1 mol dm -3 NaOH to reach the end point of a titration using phenolphthalein as the indicator. Calculate the concentration of the original hydrochloric acid. Click on the correct quantity for the original concentration of HCl, then click Next. Next 2 3 4 Made up to 250 cm 3 of diluted HCl(aq) = mol 10 cm 3 of original HCl(aq) 25 cm 3 removed is 1/10 of the dilute solution 0.100 mol dm -3 NaOH(aq) 19.8 1.98 0.198 0.0198 0.000198 19.8 cm 3 of NaOH added to reach end point = 0.00198 mol of NaOH 0.00198 Mol of HCl = 0.00198 (in 25 cm 3 of dilute solution) Made up to 250 cm 3 of diluted HCl(aq) = 0.0198 mol Well done. The concentration of the original acid is 1.98 mol dm -3. You have successfully answered the question

43. Storyboarding Storyboarding gives you an overview of the whole question – you readily pick up on details like the sample (aliquot) used for titration is simply 1/10 of the whole. Next

44. Question 2 Storyboarding The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Next

45. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the piece of apparatus shown on the right that should go in the space marked by A, then click Next. B A C Next

46. Apparatus feedback A To take out an exact volume like 75 cm 3 of the solution we would use a 25 cm 3 pipette three times. Back

47. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the piece of apparatus shown on the right that should go in the space marked by A, then click Next. B A C Next

48. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the piece of apparatus shown on the right that should go in the space marked by B, then click Next. B C Next

49. Apparatus feedback B We are adding the acid to the impure magnesium carbonate. In this case impurities are shown as black dots in the yellow solid. Back

50. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the piece of apparatus shown on the right that should go in the space marked by B, then click Next. B C Next

51. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the piece of apparatus shown on the right that should go in the space marked by C, then click Next. C Next

52. Apparatus feedback C We use a burette to add the volume of NaOH that exactly neutralises the remaining HCl. Back

53. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the piece of apparatus shown on the right that should go in the space marked by C, then click Next. C Next

54. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the label shown on the right that would be best to go in the space marked by 1, then click Next. 1 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) All MgCO 3 reacted, Xs HCl(aq) and impurities left Impure solid, mass of MgCO 3 + impurities = 0.600 g 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 2 3 4 5 6 Next

55. Label feedback 1 Here we are labelling the original quantity of acid. The label should include all the details needed for the calculation. Back

56. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the label shown on the right that would be best to go in the space marked by 1, then click Next. 1 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) All MgCO 3 reacted, Xs HCl(aq) and impurities left Impure solid, mass of MgCO 3 + impurities = 0.600 g 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 2 3 4 5 6 Next

57. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the label shown on the right that would be best to go in the space marked by 2, then click Next. 75 cm 3 of HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 2 3 4 5 6 All MgCO 3 reacted, Xs HCl(aq) and impurities left 0.200 mol dm -3 HCl(aq) Next

58. Label feedback 2 Here we are taking 75 cm 3 of the HCl. Notice that the labels should contain the quantities. Back

59. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the label shown on the right that would be best to go in the space marked by 2, then click Next. 75 cm 3 of HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 2 3 4 5 6 Next All MgCO 3 reacted, Xs HCl(aq) and impurities left 0.200 mol dm -3 HCl(aq)

60. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the label shown on the right that would be best to go in the space marked by 3, then click Next. Impure solid, mass of MgCO 3 + impurities = 0.600 g 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 2 3 4 5 6 Next 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) All MgCO 3 reacted, Xs HCl(aq) and impurities left

61. Label feedback 3 Here we are labelling the impure sample of magnesium carbonate. Back

62. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the label shown on the right that would be best to go in the space marked by 3, then click Next. Impure solid, mass of MgCO 3 + impurities = 0.600 g 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 3 4 5 6 Next 0.200 mol dm -3 HCl(aq) All MgCO 3 reacted, Xs HCl(aq) and impurities left 75 cm 3 of HCl(aq)

63. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the label shown on the right that would be best to go in the space marked by 4, then click Next. 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 4 5 6 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g All MgCO 3 reacted, Xs HCl(aq) and impurities left Next

64. Label feedback 4 Here we are labelling the remaining HCl once all the magnesium carbonate has reacted. Back

65. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the label shown on the right that would be best to go in the space marked by 4, then click Next. 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 4 5 6 Next 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g All MgCO 3 reacted, Xs HCl(aq) and impurities left

66. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the label shown on the right that would be best to go in the space marked by 5, then click Next. 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 5 6 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g All MgCO 3 reacted, Xs HCl(aq) and impurities left Next

67. Label feedback 5 Here we are labelling the concentration of NaOH. We would put the titre value on the arrow label beside the burette. Back

68. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the label shown on the right that would be best to go in the space marked by 5, then click Next. 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 5 6 Next 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g All MgCO 3 reacted, Xs HCl(aq) and impurities left

69. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the number of moles of NaOH added to reach the end point, then click Next. 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g All MgCO 3 reacted, Xs HCl(aq) and impurities left 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 17.8 x 0.200 0.075 x 0.200 24.3 +12 + 3 x 16 0.6 / 84.3 0.0178 x 0.200 0.00356 x 1 0.01144 x 84.3 0.015 - 0.00356 Next

70. Moles of NaOH feedback To calculate the number of moles of NaOH use moles = concentration x volume (in dm 3 ). Back

71. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the number of moles of NaOH added to reach the end point, then click Next. Next 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g All MgCO 3 reacted, Xs HCl(aq) and impurities left 0.200 mol dm -3 NaOH(aq) 17.8 cm 3 of NaOH added to reach end point 24.3 +12 + 3 x 16 0.0178 x 0.200 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol 17.8 x 0.200 0.075 x 0.200 0.6 / 84.3 0.00356 x 1 0.01144 x 84.3 0.015 - 0.00356

72. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the number of moles of HCl remaining after the MgCO 3 has reacted, then click Next. 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g All MgCO 3 reacted, Xs HCl(aq) and impurities left 0.200 mol dm -3 NaOH(aq) 24.3 +12 + 3 x 16 17.8 x 0.200 0.075 x 0.200 0.6 / 84.3 0.015 - 0.00356 0.00356 x 1 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol (0.09644 / 0.6) x 100 Next

73. Moles of HCl feedback 1 To calculate the number of moles of HCl use the mole ratio and the number of moles of NaOH. The equation is: HCl + NaOH  NaCl + H 2 O Back

74. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the number of moles of HCl remaining after the MgCO 3 has reacted, then click Next. Next 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g 0.200 mol dm -3 NaOH(aq) 24.3 +12 + 3 x 16 17.8 x 0.200 0.075 x 0.200 0.6 / 84.3 0.015 - 0.00356 0.00356 x 1 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol (0.09644 / 0.6) x 100

75. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the number of moles of HCl in the original 75 cm 3, then click Next. 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g 0.200 mol dm -3 NaOH(aq) 24.3 +12 + 3 x 16 (0.09644 / 0.6) x 100 17.8 x 0.200 0.075 x 0.200 0.6 / 84.3 0.015 - 0.00356 0.01144 / 2 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol Next

76. Moles of HCl feedback 2 To calculate the number of moles of HCl use moles = concentration x volume (in dm 3 ). Back

77. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the number of moles of HCl in the original 75 cm 3, then click Next. Next 0.200 mol dm -3 HCl(aq) 75 cm 3 of HCl (aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g 0.200 mol dm -3 NaOH(aq) 24.3 +12 + 3 x 16 (0.09644 / 0.6) x 100 17.8 x 0.200 0.075 x 0.200 0.6 / 84.3 0.015 - 0.00356 0.01144 / 2 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol 75 cm 3 of HCl(aq) = 0.015 mol

78. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the number of moles of HCl that reacted with MgCO 3, then click Next. 0.200 mol dm -3 HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g 0.200 mol dm -3 NaOH(aq) 24.3 +12 + 3 x 16 (0.09644 / 0.6) x 100 0.01144 x 2 0.00572 x 84.3 0.6 / 84.3 0.015 - 0.00356 0.01144 / 2 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol 75 cm 3 of HCl(aq) = 0.015 mol HCl reacted with MgCO 3 = mol Next

79. Moles of HCl feedback 3 To calculate the number of moles of HCl used up in the reaction with MgCO 3 you need to calculate the difference between the original number of moles from the 75 cm 3 of HCl and the HCl left over after the reaction has finished. Back

80. HCl reacted with MgCO 3 = mol Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the number of moles of HCl that reacted with MgCO 3, then click Next. Next 0.200 mol dm -3 HCl(aq) Impure solid, mass of MgCO 3 + impurities = 0.600 g 0.200 mol dm -3 NaOH(aq) 24.3 +12 + 3 x 16 (0.09644 / 0.6) x 100 0.01144 x 2 0.00572 x 84.3 0.6 / 84.3 0.01144 / 2 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol 75 cm 3 of HCl(aq) = 0.015 mol 0.015 - 0.00356 HCl reacted with MgCO 3 = 0.01144 mol

81. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the number of moles of MgCO 3, then click Next. Next 0.200 mol dm -3 HCl(aq) Impure solid (0.600 g) mol of MgCO 3 = 0.200 mol dm -3 NaOH(aq) 24.3 +12 + 3 x 16 (0.09644 / 0.6) x 100 0.01144 x 2 0.00572 x 84.3 0.6 / 84.3 0.01144 / 2 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol 75 cm 3 of HCl(aq) = 0.015 mol 0.482 / 0.6 x 100 HCl reacted with MgCO 3 = 0.01144 mol

82. Moles of MgCO 3 feedback 1 To calculate the number of moles of MgCO 3 you need to use the mole ratio from the balanced equation and the number of moles of HCl. Back

83. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the number of moles of MgCO 3, then click Next. Next 0.200 mol dm -3 HCl(aq) impure solid (0.600 g) mol of MgCO 3 = 0.200 mol dm -3 NaOH(aq) 24.3 +12 + 3 x 16 (0.09644 / 0.6) x 100 0.01144 x 2 0.00572 x 84.3 0.6 / 84.3 0.01144 / 2 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol 75 cm 3 of HCl(aq) = 0.015 mol 0.482 / 0.6 x 100 HCl reacted with MgCO 3 = 0.01144 mol Impure solid (0.600 g) mol of MgCO 3 = 0.00572

84. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the M r of MgCO 3 (RAM Mg = 24.3, C = 12.0 and O =16.0), then click Next. 0.200 mol dm -3 HCl(aq) 0.200 mol dm -3 NaOH(aq) 24.3 +12 + (3 x 16) (0.09644/ 0.6) x 100 0.6 0.00572 x 84.3 0.6 / 84.3 24.3 +12+ 16 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol 75 cm 3 of HCl(aq) = 0.015 mol 0.482 / 0.6 x 100 HCl reacted with MgCO 3 = 0.01144 mol Impure solid (0.600 g) mol of MgCO 3 = 0.00572 M r of MgCO 3 = Next

85. M r of MgCO 3 feedback To calculate the relative formula mass of MgCO 3 you need to add together the relative atomic masses of all the atoms in the formula. Back

86. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the M r of MgCO 3 (RAM Mg = 24.3, C = 12.0 and O =16.0), then click Next. Next 0.200 mol dm -3 HCl(aq) 0.200 mol dm -3 NaOH(aq) (0.09644 / 0.6) x 100 0.6 0.00572 x 84.3 0.6 / 84.3 24.3 + 12 + 16 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol 75 cm 3 of HCl(aq) = 0.015 mol 0.482 / 0.6 x 100 HCl reacted with MgCO 3 = 0.01144 mol Impure solid (0.600 g) mol of MgCO 3 = 0.00572 M r of MgCO 3 = 24.3 +12 + (3 x 16) M r of MgCO 3 = 84.3

87. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the mass of MgCO 3, then click Next. Next 0.200 mol dm -3 HCl(aq) 0.200 mol dm -3 NaOH(aq) (0.09644 / 0.6) x 100 0.6 0.00572 x 84.3 0.00572 / 84.3 0.482 / 0.6 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol 75 cm 3 of HCl(aq) = 0.015 mol (0.482 / 0.6) x 100 HCl reacted with MgCO 3 = 0.01144 mol Impure solid (0.600 g) mol of MgCO 3 = 0.00572 (0.6 / 0.482) x 100 M r of MgCO 3 = 84.3 Mass of MgCO 3 =

88. Mass of MgCO 3 feedback To calculate the mass of MgCO 3 you need to multiply the number of moles by the relative formula mass (the molar mass). Back

89. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the mass of MgCO 3, then click Next. Next 0.200 mol dm -3 HCl(aq) 0.200 mol dm -3 NaOH(aq) (0.09644 / 0.6) x 100 0.6 0.00572 / 84.3 0.482 / 0.6 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol 75 cm 3 of HCl(aq) = 0.015 mol (0.482 / 0.6) x 100 HCl reacted with MgCO 3 = 0.01144 mol Impure solid (0.600 g) mol of MgCO 3 = 0.00572 (0.6 / 0.482) x100 M r of MgCO 3 = 84.3 Mass of MgCO 3 = 0.00572 x 84.3 Mass of MgCO 3 = 0.482 g

90. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the percentage by mass of MgCO 3 in the sample, then click Next. Next 0.200 mol dm -3 HCl(aq) 0.200 mol dm -3 NaOH(aq) (0.09644 / 0.6) x 100 0.6 0.00572 / 84.3 0.482 / 0.6 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol 75 cm 3 of HCl(aq) = 0.015 mol (0.482 / 0.6) x 100 HCl reacted with MgCO 3 = 0.01144 mol Impure solid (0.600 g) mol of MgCO 3 = 0.00572 (0.6 / 0.482) x 100 M r of MgCO 3 = 84.3 0.00572 x 84.3 Mass of MgCO 3 = 0.482 g Percentage by mass of MgCO 3 in the sample =

91. Percentage by mass of MgCO 3 in the sample feedback To calculate the percentage by mass of MgCO 3 you need to divide the mass of MgCO 3 by the mass of impure sample and multiply by 100. Back

92. Q2. The equation for the reaction between magnesium carbonate and hydrochloric acid is: MgCO 3 + 2HCl  MgCl 2 + H 2 O + CO 2 When 75.0 cm 3 of 0.200 mol dm -3 HCl were added to 0.600 g of impure MgCO 3 some acid was left unreacted. The unreacted acid required 17.8 cm 3 of a 0.200 mol dm -3 solution of sodium hydroxide for complete reaction. Calculate the percentage by mass of MgCO 3 in the sample. Click on the correct calculation for the percentage by mass of MgCO 3 in the sample, then click Next. Next 0.200 mol dm -3 HCl(aq) 0.200 mol dm -3 NaOH(aq) (0.09644 / 0.6) x 100 0.6 0.00572 / 84.3 0.482 / 0.6 0.01144 x 84.3 17.8 cm 3 of NaOH added to reach end point = 0.00356 mol Xs HCl(aq) = 0.00356 mol 75 cm 3 of HCl(aq) = 0.015 mol HCl reacted with MgCO 3 = 0.01144 mol Impure solid (0.600 g) mol of MgCO 3 = 0.00572 (0.6 / 0.482) x 100 M r of MgCO 3 = 84.3 0.00572 x 84.3 Mass of MgCO 3 = 0.482 g (0.482 / 0.6) x 100 Percentage by mass of MgCO 3 in the sample = 80.3% Well done. The sample is 80.3% MgCO 3 by mass. You have successfully answered the question.

93. Congratulations. You have completed the storyboard skill training. Now in order to ensure you can do it independently there are two more questions for you to do, designing your own storyboard on paper – there are answers to check yours against. Good luck and don’t cheat! Next

94. Question 3 Storyboarding Iron(II) sulfate tablets were analysed in the following experiment. Five tablets were weighed, crushed and dissolved in 100 cm 3 of 1 M sulfuric acid. This was then filtered to remove insoluble bits and the filtrate made up to 250 cm 3. 25 cm 3 portions were removed and added to 25 cm 3 of 1 M sulfuric acid and then titrated against 0.00100 M potassium manganate(VII). The average titre value was 23.50 cm 3. Calculate the mass of iron(II) in each tablet. A r of Fe = 55.8. Answer

95. Q3. Iron(II) sulfate tablets were analysed in the following experiment. Five tablets were weighed, crushed and dissolved in 100 cm 3 of 1 M sulfuric acid. This was then filtered to remove insoluble bits and the filtrate made up to 250 cm 3. 25 cm 3 portions were removed and added to 25 cm 3 of 1 M sulfuric acid and then titrated against 0.00100 M potassium manganate(VII). The average titre value was 23.50 cm 3. Calculate the mass of iron(II) in each tablet. A r of Fe = 55.8. 1. 23.50 cm 3 of 0.001 M KMnO 4 = 2.35 x 10 -5 mol 2. Ratio MnO 4 - : Fe 2+ 1 : 5 3. Moles of Fe 2+ in 25 cm 3 = 1.175 x 10 -4 4. Moles of Fe 2+ in 250 cm 3 = 1.175 x 10 -3 1/10 5. Moles of Fe 2+ in 5 tablets = 1.175 x 10 -3 6. Moles of Fe 2+ in 1 tablet = 2.35 x 10 -4 7. Mass of Fe 2+ in 1 tablet = 2.35 x 10 -4 x 55.8 = 0.0131 g Next

96. Question 4 Storyboarding A solution of iodine in hydrocarbon solvent was shaken with water until equilibrium was reached. The mixture was allowed to settle and then 5 cm 3 of the hydrocarbon solvent layer and 50 cm 3 of the aqueous layer were removed and titrated separately with sodium thiosulfate solution of concentration 0.0100 mol dm -3. The hydrocarbon 5 cm 3 aliquot required 72.0 cm 3 and the aqueous aliquot required 8.35 cm 3 of the Na 2 SO 3 (aq).The expression equilibrium constant K c for this equilibrium is [I 2 (hydrocarbon)]/[I 2 (aq)]. Calculate K c. Answer

97. Q4. A solution of iodine in hydrocarbon solvent was shaken with water until equilibrium was reached. The mixture was allowed to settle and then 5 cm 3 of the hydrocarbon solvent layer and 50 cm 3 of the aqueous layer were removed and titrated separately with sodium thiosulfate solution of concentration 0.0100 mol dm -3. The hydrocarbon 5 cm 3 aliquot required 72.0 cm 3 and the aqueous aliquot required 8.35 cm 3 of the Na 2 SO 3 (aq).The expression equilibrium constant K c for this equilibrium is [I 2 (hydrocarbon)]/[I 2 (aq)]. Calculate K c. 5 cm 3 of iodine in hydrocarbon 50 cm 3 of iodine in aqueous layer Needed 72.0 cm 3 of thiosulfate Needed 8.35 cm 3 of thiosulfate 1. Ratio of moles 72.0 : 8.35 2. Ratio of volumes 1 : 10 3. Ratio of conc = 72.0 x 10 : 8.35 x 1 4. K c = [I 2 (hydrocarbon)]/[I 2 (aq)] = 720 / 8.35 = 86.2 (no units) Next

98. Storyboarding end Well done. You have completed the storyboarding skill training. This is a useful skill to use when you have a calculation based on a detailed experiment with more than one step. You draw out the rough method and then label it and work through the calculation on the diagram or you might use a table.