3. Yield: the amount of product Theoretical yield: the amount of product we expect, based on stoichiometric calculations Actual yield: amount of product from a procedure or experiment (this is given in the question) Percent yield: x 100% actual yield theoretical yield

4. Sample problem Q - What is the % yield of H 2 O if 138 g H 2 O is produced from 16 g H 2 and excess O 2 ? Step 1: write the balanced chemical equation 2H 2 + O 2  2H 2 O Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 2 mol H 2 O 2 mol H 2 16 g H 2 143 g H 2 O = 18.02 g H 2 O 1 mol H 2 O 1 mol H 2 2.02 g H 2 Step 3: Calculate % yield 138 g H 2 O 143 g H 2 O = % yield = x 100% 96.7%= actual theoretical x 100%

5. Practice problem Q - What is the % yield of NH 3 if 40.5 g NH 3 is produced from 20.0 mol H 2 and excess N 2 ? Step 1: write the balanced chemical equation N 2 + 3H 2  2NH 3 Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 2 mol NH 3 3 mol H 2 20.0 mol H 2 227 g NH 3 = 17.04 g NH 3 1 mol NH 3 Step 3: Calculate % yield 40.5 g NH 3 227 g NH 3 = % yield = x 100% 17.8%= actual theoretical x 100%

6. Practice Problems 1.) The electrolysis of water forms H 2 and O 2. 2H 2 O  2H 2 + O 2 What is the % yield of O 2 if 12.3 g of O 2 is produced from the decomposition 14 g of H 2 O? Actual yield is 12.3 g of O 2 2.) 2KClO 3  2KCI + 3O 2 107 g of oxygen is produced by heating 300 grams of potassium chlorate (KClO 3 ). Calculate % yield. Actual yield is 107 g of oxygen

7. Practice Problems continued 3.) Fe + S  FeS –What is the % yield of ferrous sulfide (FeS) if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulfide? –What is the actual yield given? Solve for % yield.