Advanced Stoichiometry

1. Advanced Stoichiometry Chapter 9, Section 2 (pages 312 – 318) Problem Set E (p. 314 # 1 – 3) Problem Set F (p. 317 # 1 – 3)

2. Terms to Know and Understand: • Limiting Reactant – the substance that controls the quantity of product that can form in a chemical reaction. • (May not be the reactant having the lowest mass!) • Identified through stoichiometry. • Excess Reactant – the substance that is not used up completely in a reaction. • Theoretical Yield – the maximum quantity of product that a reaction could theoretically make if everything about the reaction works perfectly. The theoretical yield is found by stoichiometry. • Actual Yield – the quantity of product actually produced in a reaction. • The actual yield is found experimentally. • Usually less than the theoretical yield. • Percentage Yield – the ratio relating the actual yield of a reaction to its theoretical yield.

3. Identifying a limiting reactant: Example: A chemical reaction occurs between copper (II) oxide and hydrogen according to the following balanced equation: CuO (s) + H2 (g)  Cu (s) + H2O (g) • What is the limiting reactant when 19.9-g of CuO react with 2.02-g H2? • What is the theoretical yield of copper? • First read through other questions in the problem to see if you will be asked to find the theoretical yield of one of the products – if this is the case, do stoichiometry using this product as your ‘find’. • If you are not asked to find the mass of one of the products, choose one as your ‘find’. • Do the stoichiometrytwice using each reactant as your ‘given’. To do the remaining work I will use Cu (s) as my find’.

4. Identifying a limiting reactant: 19.9 g CuO 0.250 mol CuO • Using CuO as the given: 1 mol CuO 79.545 g CuO Step 1 0.250 mol CuO 1 mol Cu 1 mol CuO 0.250 mol Cu Step 2 0.250 mol Cu 63.546 g Cu 1 mol Cu 15.9 g Cu Step 3 Mass of copper that would be produced if CuO was the limiting reactant.

5. Identifying a limiting reactant: • Using H2 as the given: 2.02 g H2 1.00 mol H2 1 mol H2 2.0158 g H2 Step 1 1.00 mol H2 1 mol Cu 1 mol H2 1.00 mol Cu Step 2 1.00 mol Cu 63.546 g Cu 1 mol Cu 63.5 g Cu Step 3 Mass of copper that would be produced if H2 was the limiting reactant.

6. Identifying a limiting reactant: • Compare the calculated quantities of product: • When 15.9-g of Cu are produced, the 19.9-g of CuO will be completely consumed (used up). • If 63.5-g of Cu were produced, 2.02 g of H2 would be completely consumed, but there is not enough CuO to do this. • The limiting reactant is CuO, because it will be gone after 15.9-g of Cu is produced, and the reaction will stop. • The excess reactant is H2, because the amount available would allow the reaction to produce 63.5-g Cu, but the reaction stops when the limiting reactant is used up. • The theoretical yield of Cu will be 15.9-g, because the reaction will stop as soon as this amount of product is formed.

7. Identifying a limiting reactant: • Note that CuO is the limiting reactant despite the fact that there was a lesser mass of H2 available to react – it is the stoichiometric quantitythat is important! • If you are only asked to find the theoretical yield of a product you must still identify the limiting reactant, as it will determine the amount of product formed! • You can recognize limiting reactant problems because you are given amounts of two of the reactants in the problem.

8. Determining Percentage Yield actual yield theoretical yield x 100 percentage yield = For example, say we conduct the reaction between CuO and H2 with the amounts given in the example problem on slide 3. While we expect to produce 15.9-g of Cu (the theoretical yield), we find that we are only able to collect 13.8-g of Cu following the reaction. We can calculate the percentage yield as follows: percentage yield = 13.8-g 15.9-g x 100 percentage yield = 86.8%

9. What mass of silver sulfide, Ag2S, can be made from 123g of H2S obtained from a rotten egg? 4Ag + 2H2S +O2 2Ag2S + 2H2O

10. What mass, in grams, of water is produced when 80 liters of hydrogen gas react with oxygen? 2H2 + O2 2H2O

11. •       Practice Box E (p. 314 # 1 – 3) • •       Chapter 9 Review, p. 331 # 31, 32, 33

12. Determining Percentage Yield • A typical problem will give you the actual yield, but require that you first identify the limiting reactant, and use stoichiometry to calculate the theoretical yield: Calculate the percentage yield of H3PO4 if 126.2-g are recovered when 100.0-g of P4O10 react with 200.0-g H2O according to the following balanced equation: P4O10 + 6 H2O  4 H3PO4 • Identify the information given to us by the problem: • actual yield of H3PO4 is 126.2-g • available amounts of reactants: • 100.0-g of P4O10 • 200.0-g H2O • Determine what we need to calculate first • the limiting reactant • the theoretical yield of H3PO4

13. Determining Percentage Yield – Find the Limiting Reactant and Theoretical Yield limiting reactant • Using P4O10 as the given: • Using H2O as the given: 1 mol P4O10 283.882 g P4O10 100.0 g P4O10 0.3522 mol P4O10 Step 1 4 mol H3PO4 1 mol P4O10 1.409 mol H3PO4 0.3522 mol P4O10 Step 2 theoretical yield 1.409 mol H3PO4 97.9927 g H3PO4 1 mol H3PO4 138.1 g H3PO4 Step 3 200.0 g H2O 11.10 mol H2O 1 mol H2O 18.0148 g H2O Step 1 4 mol H3PO4 6 mol H2O 7.400 mol H3PO4 11.10 mol H2O Step 2 7.400 mol H3PO4 97.9927 g H3PO4 1 mol H3PO4 725.1 g H3PO4 Step 3

14. Determining Percentage Yield percentage yield = actual yield theoretical yield x 100 percentage yield = 126.2 g 138.1 g x 100 percentage yield = 91.38%