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Stoichiometry

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. 1 molKMnO 4. 1 molKCl.

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Stoichiometry

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

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

  3. 1 molKMnO4 1 molKCl KMnO4 + HCl + H2S KCl + MnCl2 + S + H2O 2 mol KMnO4 2mol KMnO4 2 mol KMnO4 8 mol H2O 5 mol S 5 mol H2S 5 moleS 2 mole KMnO4 Mole-Ratio 2 6 5 2 2 5 8 How many moles of S can be obtained from 1.5 mole KMnO4 ? 1.5 mole KMnO4 x = 3.8 Mole S

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

  5. Limiting Reactant

  6. 1 mol Fe 1 mol Fe3O4 55.85 g Fe 3 mol Fe 3Fe(s) + 4H2O(g) Fe3O4 + 4H2 1 mol H2O 1 mol Fe3O4 18.02 g H2O 4 mol H2O Limiting Reactant Calculation 16.8 g 10.0 g 1. Calculate amount of product formed by each reactant .100 mol Fe3O4 .100 mol Fe3O4 16.8 g Fe 16.8 g Fe x x = yield of product .139 mol Fe3O4 10.0 g H2O x x = 2. The Limiting reactant gives the least amount of product.

  7. 16.8 g 10.0 g 1 mol Fe 55.85 g Fe 4 mol H2O 18.02 g H2O 3 mol Fe 1 mol H2O 3Fe(s) + 4H2O(g) Fe3O4 + 4H2 Excess Reactant Calculation 3. Calculate the excess amount by subtracting the reacted amount from the starting quantity 16.8 g Fe x x x = 7.23g H2O Excess water: 10.0 g - 7.2 g = 2.7 g unreacted water

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

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