ࡱ> )+( Cbjbj >jjClL<P,,,,<>>>>>>$ bbq,,wqqqj,,<q<qHq0,D X$ qZ0 0,cc0q KEY Chemistry: Percent Yield Directions: Solve each of the following problems. Show your work, including proper units, to earn full credit. 1. Slaked lime, Ca(OH)2, is produced when water reacts with quick lime, CaO. If you start with 2 400 g of quick lime, add excess water, and produce 2 060 g of slaked lime, what is the percent yield of the reaction? 2060 g = actual yield CaO + H2O ( Ca(OH) 2 2400 g excess x g = theoretical yield  EMBED Equation.3   EMBED Equation.3  2. Some underwater welding is done via the thermite reaction, in which rust (Fe2O3) reacts with aluminum to produce iron and aluminum oxide (Al2O3). In one such reaction, 258 g of aluminum and excess rust produced 464 g of iron. What was the percent yield of the reaction? 464 g = actual yield Fe2O3 + 2 Al ( Al2O3 + 2 Fe excess 258 g x g = theoretical yield  EMBED Equation.3   EMBED Equation.3  3. Use the balanced equation to find out how many liters of sulfur dioxide are actually produced at STP if 1.5 x 1027 molecules of zinc sulfide are reacted with excess oxygen and the percent yield is 75%. 2 ZnS(s) + 3 O2(g) ( 2 ZnO(s) + 2 SO2(g) 1.5x1027 molecules excess x L = theoretical yield  EMBED Equation.3   EMBED Equation.3  4. The Haber process is the conversion of nitrogen and hydrogen at high pressure into ammonia, as follows: 700 g = actual yield N2(g) + 3 H2(g) ( 2 NH3(g) x g excess x g = theoretical yield If you must produce 700 g of ammonia, what mass of nitrogen should you use in the reaction, assuming that the percent yield of this reaction is 70%?  EMBED Equation.3   EMBED Equation.3  Answers: 1. 65% 2. 87% 3. 4.19 x 104 L SO2 4. 824 g N2 Name: ________________________ Hour: ____ Date: ___________ Chemistry: Energy and Stoichiometry Directions: Solve each of the following problems. Show your work, including proper units, to earn full credit. 1. The combustion of propane (C3H8) produces 248 kJ of energy per mole of propane burned. How much heat energy will be released when 1 000 dm3 of propane are burned at STP? 2. Carbon monoxide burns in air to produce carbon dioxide according to the following balanced equation: 2 CO(g) + O2(g) ( 2 CO2(g) + 566 kJ How many grams of carbon monoxide are needed to yield 185 kJ of energy? 3. Nitrogen gas combines with oxygen gas according to the following balanced equation: N2(g) + 2 O2(g) + 67.8 kJ ( 2 NO2(g) Assuming that you have excess nitrogen, how much heat energy must be added to 540 g of oxygen in order to use up all of that oxygen? 4. Ethyl alcohol burns according to the following balanced equation: C2H5OH(l) + 3 O2(g) ( 2 CO2(g) + 3 H2O(g) + 1 364 kJ How many molecules of water are produced if 5 000 kJ of heat energy are released? Answers: 1. 11 071 kJ 2. 18.3 g CO 3. 572 kJ 4. 6.62 x 1024 molecules H2O KEY Chemistry: Energy and Stoichiometry Directions: Solve each of the following problems. Show your work, including proper units, to earn full credit. 1. The combustion of propane (C3H8) produces 248 kJ of energy per mole of propane burned. How much heat energy will be released when 1 000 dm3 of propane are burned at STP? C3H8 + 5 O2 ( 3 CO2 + 4 H2O + heat 1000 dm3 excess x kJ  EMBED Equation.3  2. Carbon monoxide burns in air to produce carbon dioxide according to the following balanced equation: 2 CO(g) + O2(g) ( 2 CO2(g) + 566 kJ How many grams of carbon monoxide are needed to yield 185 kJ of energy? 2 CO + O2 ( 2CO2 + 566 kJ x g 185 kJ  EMBED Equation.3  3. Nitrogen gas combines with oxygen gas according to the following balanced equation: N2(g) + 2 O2(g) + 67.8 kJ ( 2 NO2(g) Assuming that you have excess nitrogen, how much heat energy must be added to 540 g of oxygen in order to use up all of that oxygen? N2 + 2 O2 + 67.8 kJ ( 2 NO2 540 g x kJ  EMBED Equation.3  4. Ethyl alcohol burns according to the following balanced equation: C2H5OH(l) + 3 O2(g) ( 2 CO2(g) + 3 H2O(g) + 1364 kJ How many molecules of water are produced if 5 000 kJ of heat energy are released? C2H5OH(l) + 3 O2(g) ( 2 CO2(g) + 3 H2O(g) + 1364 kJ x molecules 5000 kJ  EMBED Equation.3  Answers: 1. 11 071 kJ 2. 18.3 g CO 3. 572 kJ 4. 6.62 x 1024 molecules H2O   '()36   !"#$%𼱧saZ CJOJQJ"jB*CJEHOJQJUphj?kC CJUVaJjB*CJOJQJUph j5CJOJQJ\5CJH*OJQJ\5CJOJQJ\B*CJOJQJph5B*CJOJQJ\phCJH*OJQJ6CJOJQJ]6>*CJOJQJ]OJQJ6] CJOJQJCJ\5CJ ()  $%=>Rs+, ^`$a$@ ^@ $0^`0a$ 0^`0$a$C%&9:;<=>Rrsuvwxzn\z"jB*CJEHOJQJUphjrkC CJUVaJjB*CJOJQJUph j5CJOJQJ\5CJH*OJQJ\5CJOJQJ\B*CJOJQJph5B*CJOJQJ\phCJH*OJQJ CJOJQJjCJEHOJQJUjkkC CJUVaJ CJOJQJjCJOJQJU$'()*+,      1 2 6 > ? 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