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Chapter 12

Chapter 12. Stoichiometry. Stoichiometry. Stoichiometry —the calculation of quantities in chemical reactions Calculating how much of each reactant is needed and how much of each product is formed. Like a recipe: you need a certain amount of ingredients to get a certain amount of product.

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Chapter 12

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  1. Chapter 12 Stoichiometry

  2. Stoichiometry • Stoichiometry—the calculation of quantities in chemical reactions • Calculating how much of each reactant is needed and how much of each product is formed. • Like a recipe: you need a certain amount of ingredients to get a certain amount of product.

  3. Chemical Equations:What can you tell? • 1N2 + 3H2 2NH3 • The coefficients in a Balanced Chemical Equation tell the # of particles of each reactant & product • 1 molecule of N2 reacts with 3 molecules of H2 to produce 2 molecules of NH3. • OR The coefficients can represent the # of moles of each reactant & product. • 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

  4. 1 Mole 1 Mole 3 Steps of Stoichiometry • Convert to moles (from particles, mass, or volume). • Do a mole to mole conversion using the balanced chemical equation • Convert moles to desired unit (particles, mass, or volume). 1. 2. 3. BCE

  5. 1 Mole 1 Mole Stoichiometry Example • 1N2 + 3H2 2NH3 • If 2.61 L of H2gas reacts with the nitrogen in the air, how many grams of NH3 will be produced? Given BCE 1 mol H2_ 22.4 L H2 17.0 g NH3_ 1 mol NH3 2 mol NH3 3 mol H2 2.61 L H2 x x x = 1.32 g NH3

  6. Percent Yield • In a lab, the amount of product formed is often less than expected based on the balanced chemical equation. • Theoretical Yield—the amount of product that should be formed (from stoichiometry problem) • Actual Yield—the amount of product actually formed in a lab (found in an experiment). • Percent Yield—the ratio of actual to theoretical yield. • Percent Yield = actual yield x 100 theoretical yield

  7. Example • 24.8g calcium carbonate is decomposed by heating. 13.1g CaOis actually produced. What is the percent yield? • CaCO3 CaO + CO2 • Actual Yield: 13.1g CaO • Theoretical Yield: (use stoichiometry) 24.8g CaCO3 x 1 mol CaCO3x 1 mol CaOx 56.1 g CaO 100.1 g CaCO3 1 mol CaCO3 1 mol CaO • =13.9 g CaO • % Yield = 13.1 x100 13.9

  8. Limiting Reagents • Limiting Reagent—limits or determines the amount of product that can be formed in a reaction. • Excess Reagent—the reactant that is not completely used up in a reaction. Limiting Reactants video

  9. Hydrogenis limiting Oxygenis excess Hydrogenis limiting Nitrogenis excess 5N2 and 9H2 6NH3 and 2N2

  10. Solving for a Limiting Reagent: • Use stoichiometry to convert each reactant (individually) into product. • The reactant that gives you the least product is the limiting reagent. • The reactant that gives the most product is the excess reagent.

  11. Example: • Nickel replaces silver from silver nitrate in solution according to the following equation: 2AgNO3 + Ni → 2Ag + Ni(NO3)2 If you have 22.9 g of Ni and 112 g of AgNO3 ,whatmass of nickel(II) nitrate would be produced? 22.9g Ni x 1 mol Ni x 1 mol Ni(NO3)2x 182.7g Ni(NO3)2 = 71.3 g Ni(NO3)2 58.7g Ni 1 mol Ni 1 mol Ni(NO3)2 112 g AgNO3x1 mol AgNO3 x 1 mol Ni(NO3)2x 182.7g Ni(NO3)2=60.2g Ni(NO3)2 169.9g AgNO3 2 mol AgNO3 1 mol Ni(NO3)2 • AgNO3 is limiting & 60.2 g Ni(NO3)2 can be produced

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