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

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.

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 stoichiometry twice using each reactant as your ‘given’. To do the remaining work I will use Cu (s) as my find’.

Identifying a limiting reactant: Using CuO as the given: 19.9 g CuO 0.250 mol CuO 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.

Using H2 as the given: Identifying a limiting reactant: 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.

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.

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 quantity that 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.

Determining Percentage Yield actual yield theoretical yield x 100 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%

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

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

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

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

Determining Percentage Yield – Find the Limiting Reactant and Theoretical Yield 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 O 200.0 g H2O 1 mol H2O 18.0148 g H2O 11.10 mol H2O Step 1 4 mol H3PO4 6 mol H2O 11.10 mol H2O 7.400 mol H3PO4 Step 2 Step 3 7.400 mol H3PO4 97.9927 g H3PO4 1 mol H3PO4 725.1 g H3PO4

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