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Stoichiometry The study of quantitative relationships among amounts of reactants and products Must examine mole ratios See steps on page 363
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Example: Mass-Mass (Stoichiometry) How many grams of silver chloride can be produced from the reaction of 17.0 g silver nitrate with excess sodium chloride solution? 17.0 g AgNO 3 1 mole AgNO 3 1 mole AgCl 143.32 g AgCl 169.88 g AgNO 3 1 mole AgNO 3 1mole AgCl = 14.3 g AgCl
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How many grams of oxygen are required to oxidize 140. g of iron to iron (III) oxide? 4 Fe + 3O 2 2 Fe 2 O 3 140. g Fe 1 mole Fe 3 mole O 2 32.00 g O 2 55.85 g Fe 4 moles Fe 1 mole O 2 = 60.2 g O 2
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Example 3 Consider the formation of nitric acid from its elements. How many grams of oxygen will be needed to produce 75.0 grams of nitric acid? How many grams of nitrogen will be needed?
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15 buns 28 patties 17 slices of cheese ? Single cheese burgers ? Double cheese burgers
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Limiting reactants Limiting reactant – what is completely consumed and determines the amount of product Excess reactant – what is “left over” Charcoal briquettes in a grill Candle burning in a closed container
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How many grams of aluminum sulfide can form from the reaction of 9.00 g of aluminum with 8.00 g of sulfur?
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Percentage Yield % yield = actual yield from experiment X 100 theoretical yield from Stoichiometry Example: A chemist produces 3.45g of product. The maximum possible yield was calculated to be 3.97 g. What is the % yield of the process? 3.45 3. 97 X 100 = 86. 9 %
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Consider the combustion of methane. Find the percent yield if 10.0 grams of methane is burned in 45.0 grams of oxygen producing 19.5 grams of water.
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Given: NaOH + Cr(OH) 3 NaCr(OH) 4 What is your percent yield if you begin with 66.0 g of Cr(OH) 3 and obtain 38.4 g of NaCr(OH) 4 91. 6 g NaCr(OH) 4 hence 41.9% not a great % in the marketplace
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