Ch4_C_LimReact - Reactions Involving a LIMITING REACTANT...

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Unformatted text preview: Reactions Involving a LIMITING REACTANT LIMITING REACTANTS • In a given reaction, there is not enough of one reagent to use up the other reagent completely. LIMITING REACTANTS • The reagent in short supply LIMITS the quantity of product that can be formed. Reactants Products 2 NO(g) + O 2 (g) 2 NO 2(g) Limiting reactant = ___________ Excess reactant = ____________ LIMITING REACTANTS LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl2 + H2 mass Zn (g) Rxn 1 Rxn 7.00 React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl2 + H2 Rxn 2 3.27 Rxn 3 1.31 mass Zn (g) mol Zn mol HCl mol HCl/mol Zn Rxn 1 Rxn 7.00 0.107 0.100 0.93 Rxn 2 3.27 0.050 0.100 2.00 Page 1 Rxn 3 1.31 0.020 0.100 5.00 Reaction to be Studied 2 Al + 3 Cl2 ---> Al2Cl6 Step 1 of LR problem: Step 1 of LR problem: compare actual mole ratio compare actual mole ratio of reactants to theoretical of reactants to theoretical mole ratio. mole ratio. PROBLEM: Mix 5.40 g of Al with 8.10 g of PROBLEM: Mix 5.40 g of Al with 8.10 g of PROBLEM: Mix Cl22. How many grams of Al 22Cl6 can form? Cl . How many grams of Al Cl6 can form? Mass product Mass reactant Moles reactant Stoichiometric factor Moles product Deciding on the Limiting Reactant 2 Al + 3 Cl2 ---> Al2Cl6 If Step 1 of LR problem: Step 1 of LR problem: compare actual mole ratio compare actual mole ratio of reactants to theoretical of reactants to theoretical mole ratio. mole ratio. Deciding on the Limiting Reactant 2 Al + 3 Cl2 ---> Al2Cl6 If mol Cl2 3 > 2 mol Al then there is not enough Al to use up all the Cl 2, and the limiting reagent is Al mol Cl2 3 < 2 mol Al then there is not enough Cl 2 to use up all the Al, and the limiting reagent is Cl2 Page 2 2 Al + 3 Cl2 ---> Al2Cl6 Reactants must be in the mole ratio 3 mol Cl2 = mol Al 2 Step 2 of LR problem: Calculate Step 2 of LR problem: Calculate moles of each reactant moles of each reactant We have 5.40 g of Al and 8.10 g of Cl 2 5.40 g Al • 8.10 g Cl2 • 1 mol = 0.200 mol Al 27.0 g 1 mol = 0.114 mol Cl2 70.9 g Find mole ratio of reactants Find mole ratio of reactants Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al2 Cl6 can form? 2 Al + 3 Cl2 ---> Al2Cl6 2 Al + 3 Cl2 ---> Al2Cl6 mol Cl2 0.114 mol = 0.200 mol mol Al = 0.57 This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio. Limiting reagent is Limiting Cl2 Cl How much of which reactant will How much of which reactant will remain when reaction is complete? remain when reaction is complete? • Cl2 was the limiting reactant. Therefore, Al was present Limiting reactant = Cl 22 Limiting reactant = Cl Base all calcs. on Cl22 Base all calcs. on Cl grams Al2Cl6 grams Cl2 1 mol Al2Cl6 3 mol Cl2 moles Cl2 moles Al2Cl6 Calculating Excess Al Calculating Excess Al 2 Al + 3 Cl2 Al Cl 0.200 mol 0.200 mol products products 0.114 mol = LR 0.114 mol = LR CALCULATIONS: calculate mass of CALCULATIONS: calculate mass of Al22Cl 66 expected. Al Cl expected. Step 1: Calculate moles of Al 2Cl6 Step Calculate expected based on LR. 0.114 mol Cl2 • Step 2: Calculate mass of Al 2Cl6 expected Step Calculate based on LR. 0.0380 mol Al 2Cl6 • 266.4 g Al2Cl6 = 1 0.1 g Al2Cl6 mol Calculating Excess Al Calculating Excess Al 2 Al + 3 Cl2 Al Cl 0.200 mol 0.200 mol products products 0.114 mol = LR 0.114 mol = LR 0.114 mol Cl2 • in excess. But how much? 1 mol Al2Cl6 = 0 .0380 mol Al2Cl6 3 mol Cl2 2 mol Al = 0.0760 mol Al req'd 3 mol Cl2 • First find how much Al was required. Excess Al = Al available - Al required • Then find how much Al is in excess. Then is = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess 0.124 Page 3 ...
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This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Staff during the Spring '08 term at University of Central Florida.

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