1. Stoichiometry In Solution Chemistry

2. Stochiometry involves calculating the amounts of reactants and products in chemical reactions. If you know the atoms or ions in a formula or a reaction, you can use stoichiometry to determine the amounts of atoms or ions that react.

3. The Concentration of Ions Problem Calculate the concentration (in mol/L) of chloride ions in the following solutions. 19.8 g of potassium chloride dissolved in 100 mL solution 26.5 g of calcium chloride dissolved in 150 mL solution A mixture of the two solutions (volumes are additive)

4. 19.8 g of potassium chloride dissolved in 100 mL solution KCl  39.0983 + 35.4527 = 74.5510 g/mol 19.8 g KCl x 1 mol = 0.266 mol KCL 74.5510 g KCl  K + + Cl - So for every KCl mole there is a Cl - mole

5. We calculated 0.266 mol KCl so there is also 0.266 mol Cl - 0.266 mol Cl - / 0.100 L = 2.66 mol/L Cl -

6. 26.5 g of calcium chloride dissolved in 150 mL solution CaCl 2  40.078 + 2(35.4527) = 110.983 g/mol 26.5 g CaCl 2 x 1 mol = 0.239 mol CaCl 2 110.983 g CaCl 2  Ca 2+ + 2 Cl - So for every CaCl 2 mole there is 2 Cl - moles

7. We calculated 0.239 mol CaCl 2 so there is 0.478 mol Cl - 0.478 mol Cl - / 0.150 L = 3.19 mol/L Cl -

8. A mixture of the two solutions (volumes are additive) Total amount of Cl - moles is 0.266 + 0.478 = 0.744 Total volume is 0.100 + 0.150 = 0.250 L Total concentration of Cl - is 0.744 mol / 0.250 L = 2.98 mol/L

9. Mass Percent of Ions Problem Rhubard leaves contain relatively high concentrations of oxalate ions (C 2 O 4 2- ), which are poisonous and cause respiratory problems. To determine the percent of oxalate ions, a student measure a mass of leaves to be 238.6 g. Then the students ground up the leaves and added calcium chloride solutions to precipitate out the calcium oxalate. The dried calcium oxalte was 0.556 g. What was the mass percent of oxalate in the leaves?

10. Find molar mass of calcium oxalate Use the given mass of dried calcium oxalate and the molar mass to determine the number of moles Write a net ionic equation for the formation of calcium oxalate Use the coefficients to determine a ratio between calcium oxalate and oxalate Once you have moles of oxalate, find the molar mass and then determine the mass of oxalate. Using the determined mass of oxalate and the mass of the leaves, determine the mass percent of oxalate.

11. Find molar mass of calcium oxalate CaC 2 O 4  40.078 + 2(12.011) + 4(15.9994) = 128.098 Use the given mass of dried calcium oxalate and the molar mass to determine the number of moles 0.556 g x 1 mol= 0.00434 mol CaC 2 O 4 128.098g Write a net ionic equation for the formation of calcium oxalate Ca 2+ + C 2 O 4 2-  CaC 2 O 4

12. Use the coefficients to determine a ratio between calcium oxalate and oxalate One C 2 O 4 2- makes one CaC 2 O 4 0.00434 mol CaC 2 O 4 means 0.00434 mol C 2 O 4 2- Once you have moles of oxalate, find the molar mass and then determine the mass of oxalate. (2(12.011) + 4(15.9994) = 88.020 g/mol) 0.00434 mol C 2 O 4 2- x 88.020 g = 0.382 g 1 mol Using the determined mass of oxalate and the mass of the leaves, determine the mass percent of oxalate. 0.382 g/238.6 g x 100% = 0.160%

13. Finding Minimum Volume to Precipitate Aqueous solutions that contain silver ions are usually treated with chloride ions to recover silver chloride. What is the minimum volume of 0.25 mol/L magnesium chloride needed to precipitate all the silver ions in 60 mL of 0.30 mol/L silver nitrate? (Assume that silver chloride is completely insoluble in water).

14. Find moles of silver chloride from the volume and concentration. Write and balance a chemical equation for the reaction Use the mole ratio between silver chloride and magnesium chloride to determine how much magnesium chloride is needed. Use that amount and the concentration of the solution to determine the volume needed.

15. Find moles of silver chloride from the volume and concentration. 60 mL of 0.30 mol/L 0.060 L x 0.30 mol = 0.018 mol AgNO 3 1 L Write and balance a chemical equation for the reaction MgCl 2 + 2 AgNO 3  Mg(NO 3 ) 2 + 2 AgCl Use the mole ratio between silver chloride and magnesium chloride to determine how much magnesium chloride is needed 1 MgCl 2 to 2 AgNO 3 (so 0.0090 mol MgCl 2 )

16. Use that amount and the concentration of the solution to determine the volume needed. 0.0090 mol MgCl 2 0.25 mol/L 0.0090 mol = 0.036 L of MgCl 2 needed. 0.25 mol/L