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Chapter 12. Stoichiometry. Stoichiometry. STOY-KEE-AHM-EH-TREE Founded by Jeremias Richter, a German chemist Greek orgin: stoikheion: element & metron: measure Stoichiometry —the calculation of quantities in chemical reactions

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## Chapter 12

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**Chapter 12**Stoichiometry**Stoichiometry**• STOY-KEE-AHM-EH-TREE • Founded by Jeremias Richter, a German chemist • Greek orgin: stoikheion: element & metron: measure • 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.**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. • ORThe 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.**You Should Know About Stoichiometry**• You have to use units to understand the process • Units include g, mol, L, particles & chemical formulas • Convert your given to moles • You have to use units to understand the process • Use BCE coefficients to convert from one substance to another • You have to use units to understand the process • Convert answer to desired unit • You have to use units to understand the process • Units include g, mol, L, particles & chemical formulas • It really is not that hard **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**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**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**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**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**Hydrogenis**limiting Oxygenis excess Hydrogenis limiting Nitrogenis excess 5N2 and 9H2 6NH3 and 2N2**Solving for a Limiting Reagent:**• Use stoichiometry to convert each reactant (individually) into product (that’s 2 stoichiometry problems) • The reactant that gives you the least product is the limiting reagent. • The reactant that gives the most product is the excess reagent.**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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