Stoichiometry Preparation for College Chemistry Luis Avila Columbia University Department of Chemistry

The Mole-ratio method Mole-Mole Calculations Mole-Mass Calculations Mass-Mass calculations Yield calculations

Mole-Ratio 2 6 5 2 2 5 8 KMnO4 + HCl + H2S KCl + MnCl2 + S + H2O 2 6 5 2 2 5 8 KMnO4 + HCl + H2S KCl + MnCl2 + S + H2O 1 molKMnO4 1 molKCl 2mol KMnO4 5 mol S 2 mol KMnO4 5 mol H2S 2 mol KMnO4 8 mol H2O How many moles of S can be obtained from 1.5 mole KMnO4 ? 2 mole KMnO4 5 moleS 1.5 mole KMnO4 x = 3.8 Mole S

Mole-Mass Calculation 2 6 5 2 2 5 8 KMnO4 + HCl + H2S KCl + MnCl2 + S + H2O How many g of S can be obtained from 1.5 g KMnO4 ? 158.04 g KMnO4 1 mol KMnO4 2mol KMnO4 5 mol S 1mol S 32.07 g S 1.5 g KMnO4 x x x = 80.9 g S

Limiting Reactant

Limiting Reactant Calculation 3Fe(s) + 4H2O(g) Fe3O4 + 4H2 1. Calculate amount of product formed by each reactant 55.85 g Fe 1 mol Fe 3 mol Fe 1 mol Fe3O4 .100 mol Fe3O4 .100 mol Fe3O4 16.8 g Fe x 16.8 g Fe x = yield of product 18.02 g H2O 1 mol H2O 4 mol H2O 1 mol Fe3O4 .139 mol Fe3O4 10.0 g H2O x x = 2. The Limiting reactant gives the least amount of product.

Excess Reactant Calculation 3Fe(s) + 4H2O(g) Fe3O4 + 4H2 16.8 g 10.0 g 3. Calculate the excess amount by subtracting the reacted amount from the starting quantity 55.85 g Fe 1 mol Fe 3 mol Fe 4 mol H2O 18.02 g H2O 1 mol H2O 16.8 g Fe x x x = 7.23g H2O Excess water: 10.0 g - 7.2 g = 2.7 g unreacted water

Percentage Yield Calculation Theoretical Yield Actual Yield x 100 In the previous reaction the theoretical yield was 231.55 g Fe3O4 1 mol Fe3O4 23.2 g Fe3O4 .100 mol Fe3O4 x = If the actual amount obtained is 16.2 g, then the % yield: 16.2 g Fe3O4 23.2 g Fe3O4 Percentage Yield = x 100 = 69.8%