1. Stoichiometry Calculations Limiting Reactants and Percent Yield Granada Hills Charter High School Student Review and Exam Preparation

2. The Mole-ratio concepts Mole-Mole Calculations Mole-Mass Calculations Mass-Mass Calculations Percent Yield Calculations

3. 2 6 5 2 2 5 8 KMnO 4 + HCl + H 2 S KCl + MnCl 2 + S + H 2 O Mole-Ratio Concept 2 mol KMnO 4 2 mol KCl 2 mol KMnO 4 5 mol H 2 S 2 mol KMnO 4 5 mol S 2 mol KMnO 4 8 mol H 2 O How many moles of S can be obtained from 1.5 mole KMnO 4 ? 1.5 mole KMnO 4 x 2 mole KMnO 4 5 mole S = 3.8 Mole S KMnO 4 S

4. 2 6 5 2 2 5 8 Mass-Mole-Mass Calculation How many grams of S can be obtained from 1.5 g KMnO 4 ? 1.5 g KMnO 4 x = 80.9 g S 158.04 g KMnO 4 1 mol KMnO 4 x 2mol KMnO 4 5 mol S x 1mol S 32.07 g S KMnO 4 + HCl + H 2 S KCl + MnCl 2 + S + H 2 O

5. Which is the limiting reactant? Use the relationships from the balanced chemical equation Use the relationships from the balanced chemical equation You take each reactant in turn and ask how much product would be obtain, if each were totally consumed. You take each reactant in turn and ask how much product would be obtain, if each were totally consumed. The reactant that gives the smaller amount of product is the limiting reactant. The reactant that gives the smaller amount of product is the limiting reactant. Strategy:

6. Limiting Reactant Calculation 3Fe(s) + 4H 2 O(g) Fe 3 O 4 + 4H 2 16.8 g10.0 g 16.8 g Fe x x 55.85 g Fe 1 mol Fe 3 mol Fe 1 mol Fe 3 O 4 10.0 g H 2 O x x 18.02 g H 2 O 1 mol H 2 O 4 mol H 2 O 1 mol Fe 3 O 4 = =.100 mol Fe 3 O 4.139 mol Fe 3 O 4 1. Calculate amount of product formed by each reactant 2. The Limiting reactant gives the least amount of product..100 mol Fe 3 O 4 16.8 g Fe Maximum yield of product

7. How much Excess Reactant Remains? Start with the original amount of the already identified Limiting Reactant. Start with the original amount of the already identified Limiting Reactant. Use the relationships from the balanced chemical equation to find how much of the excess reactant was used by the limiting reactant to make the product. Use the relationships from the balanced chemical equation to find how much of the excess reactant was used by the limiting reactant to make the product. Subtract this quantity from the original amount of the excess reactant provided. Subtract this quantity from the original amount of the excess reactant provided. Strategy:

8. 3Fe(s) + 4H 2 O(g) Fe 3 O 4 + 4H 2 16.8 g10.0 g Excess Reactant Calculation 4. Calculate the excess amount by subtracting the reacted amount from the original starting quantity 16.8 g Fe x 55.85 g Fe 1 mol Fe 3 mol Fe 4 mol H 2 O xx 18.02 g H 2 O 1 mol H 2 O = 7.23g H 2 O Excess water: 10.0 g - 7.23 g = 2.7 g unreacted water 3. Calculate the amount of the other reactant (XS) that was utilized by the limiting reactant to form the product. L/RX SX S This amount was used to make the product

9. Percent Yield Calculations If the actual amount obtained is 16.2 g, then the % yield: Theoretical Yield Actual Yield x 100 In the previous reaction the theoretical yield was 0.100 mol Fe 3 O 4 231.55 g Fe 3 O 4 1 mol Fe 3 O 4 x = 23.2 g Fe 3 O 4 16.2 g Fe 3 O 4 23.2 g Fe 3 O 4 x 100 =69.8%Percentage Yield =