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Chemistry HS/Science Unit: 08 Lesson: 01 Limiting Reactants Practice KEY 1. How many grams of aluminum chloride could be produced from 35.5 g of aluminum and 39.0 g of chlorine gas? 2Al + 3 Cl2 35.5 g Al X • • • 2AlCl3 1 mol Al X 3 mol Cl2 X 70.906 g Cl2 = 140. g Cl2 26.982 g Al 2 mol Al 1 mol Cl2 This means 35.5 g Al would react with 140. g Cl2 Only 39.0 g Cl2 were available, so it is the limiting reactant. Use the mass of Cl2 in the reaction to determine the mass of the AlCl 3 formed. 39.0 g Cl2 = __ ?______ g AlCl3 39.0 g Cl2 X 1 mol Cl2 X 70.906 g Cl2 2 mol AlCl3 X 133.341 g AlCl 3 = 3 mol Cl2 1 mol AlCl2 Answer: 48.9 g AlCl 3 2. Identify the limiting reagent when 10.8 g water reacts with 4.7 g Na to produce NaOH and H 2. To solve, set one reactant to the other and compare what is calculated to the actual amount in the reaction to determine the theoretical limiting reactant. 10.8 g H2O = ____?__ g Na Or 4.7 g Na = ____?__ g H2O 2H2O + 2 Na 2 NaOH + H2 10.8 g H2O X 1 mol H2O x 2 mol Na 18.015g H2O 2 mol H2O • X 22.990 g Na = 13.8 g Na 1 mol Na This means that 10.8 g will react completely with 13.8 g Na, but only 4.7 g Na was available to this reaction. Therefore, Na is the limiting reactant in this problem. ©2012, TESCCC 05/08/13 page 1 of 2 Chemistry HS/Science Unit: 08 Lesson: 01 3. If 31.2 g of barium nitrate and 21.9 g of ammonium carbonate are mixed, how many moles of ammonium nitrate are produced? Ba(NO3)2 + (NH4)2 CO3 BaCO3 + 2NH4 NO3 After balancing the chemical equation, calculate the limiting reactant and then use the limiting reactant to determine the mass of the product formed. 31.2 g BaNO3 = __?____ g (NH4)2 CO3 OR 21.9 g (NH4)2CO3 = __?____g BaNO3 31.2 g BaNO3 X 1 mol BaNO3 X 1 mol (NH4)2CO3 X 96.086 g (NH4)2CO3 = 11.5 g (NH4)2CO3 199.332 g BaNO3 1 mol BaNO3 1 mol (NH4)2CO3 • • • 31.2 g BaNO3 would react with 11.5 g (NH4)2 CO3. Since 21.9 g of (NH4)2 CO3 was available, then it is the excess reactant. Therefore, BaNO3 is the limiting reactant. Use the limiting reactant to determine the mass of the product. 31.2 g BaNO3 = _________ g NH4 NO3 31.2 g BaNO3 X 1 mol BaNO3 X 2 mol NH4 NO3 199.332 g BaNO3 1 mol Ba(NO3)2 = 0.313 mol NH4 NO3 ©2012, TESCCC 05/08/13 page 2 of 2