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Ch. 9: Calculations from Chemical Equations

Ch. 9: Calculations from Chemical Equations. Stoichiometry Mole-Mole Calculations Mole-Mass Calculations Mass-Mass Calculations. Moleville. Moletown. Mole Ratio Bridge. Molar Mass Railroad. Molar Mass Railroad. Mass Junction. Mass Valley. Stoichiometry.

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Ch. 9: Calculations from Chemical Equations

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  1. Ch. 9: Calculations from Chemical Equations Stoichiometry Mole-Mole Calculations Mole-Mass Calculations Mass-Mass Calculations

  2. Moleville Moletown Mole Ratio Bridge Molar Mass Railroad Molar Mass Railroad Mass Junction Mass Valley

  3. Stoichiometry • Stoichiometry: calculations based on a balanced chemical equation Mole Ratio Moles of “A” Moles of “B” Molar Mass Molar Mass Grams of “A” Grams of “B” • Mole ratio: ratio of coefficients of any two substances in a balanced chemical equation

  4. 4 mol O2 1 x Mole-Mole Calculations • How many moles of water can be obtained from the reaction of 4 moles of O2? 2 H2 (g) +1O2 (g) → 2 H2O (g) 2 mol H2O 1 mol O2 = 8 mol H2O Mole Ratio

  5. 8 mol H2 1 x • How many moles of NH3 can be obtained from the reaction of 8 moles of H2? __ H2 (g) + __ N2 (g) → __ NH3 (g) 3 1 2 2 mol NH3 3 mol H2 = 5.33 mol NH3 Mole Ratio

  6. 6 mol Al 1 x x Mole-Mass Calculations 2 Al (s) + 6 HCl (aq) → 2 AlCl3 (aq) + 3 H2 (g) • What mass of hydrogen gas can be produced by reacting 6 moles of aluminum with HCl? 3 mol H2 2 mol Al 2.0 g H2 1 mol H2 = 18 g H2 Mole Ratio Molar Mass

  7. 6 mol Al 1 x x 2 Al (s) + 6 HCl (aq) → 2 AlCl3 (aq) + 3 H2 (g) • What mass of HCl is needed to react with 6 moles of aluminum? 6 mol HCl 2 mol Al 36.0 g HCl 1 mol HCl = 648 g HCl Mole Ratio Molar Mass

  8. Mass-Mass Calculations Sn(s) + 2 HF (g) → SnF2 (s) + H2 (g) How many grams of SnF2 can be produced from the reaction of 30.00 g of HF with Sn? 1 molSnF2 2 mol HF 156.71 g SnF2 1 mol SnF2 30.00 g HF 1 1 mole HF 20.01 g HF x x x = 117.5 g SnF2 Molar Mass Molar Mass Mole Ratio

  9. Limiting Reactant • controls the amount of product formed. CO(g) + 2H2 (g)  Ch3OH • If 500 mol of CO react with 750 mol of H2, which is the limiting reactant? • Use either given amount to calculate required amount of other. • Compare calculated amount to amount given b. How many moles of excess reactant remain unchanged? H2 125 mol CO

  10. Percent yield= (actual yield/ theoretical yield)*100 • Theoretical yield is the maximum amount of product that can be produced from a given amount of reactant • Actual yield is the measured amount of a product obtained from a reaction Theoretical yield= 117.5 g SnF2 Actual yield = 113. 4g SnF2 Percent yield = 113.4 g SnF2 117.5 g SnF2 *100

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