1. Stoichiometry Chapter 9

2. Stoichiometry: Calculation of the quantities of substances involved in chemical reactions 4 Classes of Problems: Mole-Mole Mole-Mass Mass-Mole Mass-Mass

3. Information Given In A Chemical Equation: Coefficients in a balanced equation represent ratios Example: CO + 2H 2  CH 3 OH 1:2:1 ratio 1 molecule of CO for every 2 molecules of H 2 for every 1 molecule of CH 3 OH If I want 50 molecules of CH 3 OH, how many molecules of H 2 do I need?

4. Usually ratios in an equation represent moles Example: 1 mol CO + 2 mol H 2  1 mol CH 3 OH BEFORE DETERMINING THE RATIO OF AN EQUATION, REMEMBER TO BALANCE FIRST

5. Mole Ratio: Conversion factor that relates the number of moles of any two substances involved in a balanced chemical reaction (balanced equation) Example: 2Al 2 O 3  4Al + 3O 2 How many mol of Al are produced from 13 mol of Al 2 O 3 ?

6. units you want = units you have x conversion factor Conversion Factor = Equivalence Statement Ratio from balanced equation what you want what you have

7. 2Al 2 O 3  4Al + 3O 2 Possible Ratios: 2 mol Al 2 O 3 4 mol Al 4 mol Al2 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 3 mol O 2 2 mol Al 2 O 3 4 mol Al3 mol O 2 3 mol O 2 4 mol Al

8. 2Al 2 O 3  4Al + 3O 2 Possible Ratios: 2 mol Al 2 O 3 4 mol Al 4 mol Al2 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 3 mol O 2 2 mol Al 2 O 3 4 mol Al3 mol O 2 3 mol O 2 4 mol Al

9. ? mol Al = 13 mol Al 2 O 3 x = 4 mol Al 2 mol Al 2 O 3 26 mol Al DO: Example 9.3 on Board Self Check 9.1 pg 257 in Notes Section Review 1-5 pg 258 in Notes

10. Mole-Mole Calculations: you are asked to find the number of moles of a substance produced from a given number of moles of another substance Given moles of substance A  ? moles of substance B * Conversion Factor is now a mole ratio

11. units you want = units you have x conversion factor Conversion Factor is now a mole ratio so: moles B = moles A x mole ratio YOU WANT “B” YOU HAVE “A” BABA

12. Q: How many moles of NH 3 are produced when 6 moles of H (g) react with N (g) ? 1.Write and balance equation H 2 + N 2  NH 3 2. Determine A and B A= Hydrogen B= Ammonia 3.Set up according to formula __ NH 3 = 6 mol H 2 x 4. Solve 2 mol NH 3 3 mol H 2 4 3H 2 + N 2  2NH 3 2 mol NH 3 3 mol H 2 __ mol NH 3 = 6 mol H 2 x

13. Q: Potassium chlorate decomposes into potassium chloride and oxygen gas. How many moles of potassium chlorate are needed to produce 15 moles of oxygen? 1. Write and balance equation KClO 3  KCl + O 2 2. Determine A and B A= Oxygen B= Potassium chlorate 3. Set up according to formula 4. Solve __ mol KClO 3 = 15 mol O 2 x 2KClO 3  2KCl + 3O 2 2 mol KClO 3 3 mol O 2 10

14. Mole-Mass Calculations: You are asked to find the number of grams of a substance produced from the number of moles of another substance Calculate the same way as mole-mole problems, but with one extra step: Find the molar mass molar mass = gram molecular weight

15. mass B = moles A x mole ratio x BABA molar mass B 1 mole B

16. Q: Potassium chlorate decomposes into potassium chloride and oxygen gas. How many moles of potassium chlorate are needed to produce 15 moles of oxygen? How many grams of potassium chlorate? 1.Write and balance equation KClO 3  KCl + O 2 2.Determine A and B A= Oxygen B= Potassium chlorate 3.Set up according to formula 4. Solve __ mol KClO 3 = 15 mol O 2 x 5. Find Molar Mass 10 x molar mass of KClO 3 = _______ g 2KClO 3  2KCl + 3O 2 2 mol KClO 3 3 mol O 2 10 1225.5

17. Mass-Mole Calculations: You are given the mass in grams of a substance and are asked to find the moles of the second substance moles B = mass A x x mole ratio BABA 1 mole A molar mass A

18. Example: The industrial solvent carbon disulfide (CS 2 ) and carbon monoxide (CO) are produced through a reaction between carbon (C) and sulfur dioxide (SO 2 ). If 8 grams of SO 2 react: a)How many moles of CS 2 are formed? b)How many moles of CO are formed?

19. a) 1.Write the balanced equation 5C + 2SO 2  CS 2 + 4CO 2.Find molecular mass of A (SO 2 ) 32.07 + 16 + 16 = 64.07g/mol 3.Set up to find moles B (CS 2 ) 8g SO 2 x x mole ratio 4. Do calculation = 1 mol A 64.07g 1 mol CS 2 2 mol SO 2.062 mol CS 2

20. b) 1.Write the balanced equation 5C + 2SO 2  CS 2 + 4CO 2.Find molecular mass of A (SO 2 ) 32.07 + 16 + 16 = 64.07g/mol 3.Set up to find moles B (CO) 8g SO 2 x x mole ratio 4. Do calculation = 1 mol A 64.07g 4 mol CO 2 mol SO 2.25 mol CO

21. Mass-Mass Calculations: You are given the mass in grams of one substance and are asked for the mass in grams of a second substance mass B = mass A x x mole ratio x 1 mole A molar mass A molar mass B 1 mole B BABA

22. Limiting Reagent: The substance in a chemical reaction that runs out first. Example:A + B  C (but how much C?) A = Stoppers (20) B = Test Tubes (30) C = Test Tubes with Stoppers Limiting Reagent = Stoppers Excess Reagent = Test Tubes (10) Product = 20 Test Tubes with Stoppers

23. Q:A cup of coffee costs 50 cents. In your possession you have 100g each of nickels, dimes, and quarters. 1 nickel = 5g 1 dime = 2.22g 1 quarter = 5.55g How many cups of coffee can you buy? This type of problem models what you have to do with limiting reagents in chemistry.

24. 1 nickel = 5g 100/5 = 20 x.05 = $1.00 1 dime = 2.22g 100/2.22 = 45 x.10 = $4.50 1 quarter = 5.55g = 555g 100/5.55 = 18 x.25 = $4.50 Total = $10.00 /.50 per cup = 20 cups The limiting reagent is the value of the coins

25. To Find Limiting Reagent: 1.Do mass calculations for both quantities given in the stoichiometry problem 2. Whichever calculation gives the smallest number is your limiting reagent

26. Steps to Solve Stoichiometry Problems Involving Limiting Reagents 1. Balance the Equation 2. Convert masses of reactants to moles 3. Determine limiting reactant using number of moles and the appropriate mole ratios 4. Compute the number of moles of desired product using the amount of limiting reactant and the appropriate mole ratio 5. Convert moles to mass

27. Example: 25kg of N 2 and 5kg of H 2 are reacted to form NH 3. Calculate the mass of NH 3 produced from a complete reaction. Step 1: Balance Step 2: Convert Masses to Moles 25000g N 2 x 5000g H 2 x 1mol N 2 28.02g N 2 = 892.22 mol N 2 1mol H 2 2.02g H 2 = 2475.25 mol H 2 N 2 + 3H 2  2NH 3

28. Step 3: Determine Limiting Reactant 892.22 mol N 2 x 2475.25 mol H 2 x Which One is the Limiting Reactant? Why? H 2 N 2 3 mol H 2 1 mol N 2 3 mol H 2 = 2676.66 mol H 2 = 825.08 mol N 2 WE NEED 2676.66 mol WE HAVE 2475.25 mol WE NEED 825.08 mol WE HAVE 892.22 mol

29. Step 4: Using Limiting Reactant Amount and Mole Ratios, Find the Moles of Product 2475.25 mol H 2 x Step 5: Convert Moles to Mass 2 mol NH 3 3 mol H 2 = 1650.17 mol NH 3 1650.17 mol NH 3 x 17.03g NH 3 1 mol NH 3 = 28,102.40g NH 3 28.1kg NH 3

30. Finding Limiting Reagent Quantity in Stoichiometry Problems Amount of Excess Reactant = Original Quantity of Excess Reactant - Amount of Product Predicted by Limiting Reactant Amount of Product Predicted by Excess Reactant x

31. Percent Yield Percent Yield = Actual Yield Theoretical Yield X 100