Ch4_C_LimReact - Reactions Involving a LIMITING REACTANT...

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Unformatted text preview: Reactions Involving a LIMITING REACTANT • In a given reaction, there is not enough of one reagent to use up the other reagent completely. • The reagent in short supply LIMITING REACTANTS LIMITING REACTANTS LIMITS the quantity of product the that can be formed. Reactants Products 2 NO(g) + O 2 (g) 2 NO 2(g) Limiting reactant = ___________ Excess reactant = ____________ LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) (aq) Zn + 2 HCl ---> ZnCl2 + H2 Rxn 1 Rxn 7.00 Rxn 2 3.27 Rxn 3 1.31 LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) (aq) Zn + 2 HCl ---> ZnCl2 + H2 Rxn 1 Rxn 7.00 0.107 0.100 0.93 Rxn 2 3.27 0.050 0.100 2.00 Rxn 3 1.31 0.020 0.100 5.00 Reaction to be Studied 2 Al + 3 Cl2 ---> Al2Cl6 ---> mass Zn (g) mass Zn (g) mol Zn mol HCl mol HCl/mol Zn HCl/ Page 1 PROBLEM: Mix 5.40 g of Al with 8.10 g of PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl22. How many grams of Al 22Cl6 can form? Cl . How many grams of Al Cl6 can form? can Mass reactant Mass product 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. Moles reactant Stoichiometric factor Moles product 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. 2 Al + 3 Cl2 ---> Al2Cl6 ---> Reactants must be in the mole ratio 3 mol Cl2 = mol Al 2 Deciding on the Limiting Reactant Deciding on the Limiting Reactant 2 Al + 3 Cl2 ---> Al2Cl6 ---> If 2 Al + 3 Cl2 ---> Al2Cl6 ---> If 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 mol Cl2 3 > 2 mol Al mol Cl2 3 < 2 mol Al 5.40 g Al • 1 mol = 0.200 mol Al 27.0 g 1 mol = 0.114 mol Cl2 70.9 g then there is not enough Al to use up all the Cl 2, and the limiting reagent is then there is not enough Cl 2 to to use up all the Al, and the limiting reagent is Al Cl2 8.10 g Cl2 • Page 2 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? can 2 Al + 3 Cl2 ---> Al2Cl6 ---> CALCULATIONS: calculate mass of CALCULATIONS: calculate mass of Al22Cl 66 expected. Al Cl expected. expected. Step 1: Calculate moles of Al 2Cl6 expected based on LR. 0.114 mol Cl2 • 1 mol Al2Cl6 = 0 .0380 mol Al2Cl6 3 mol Cl2 2 Al + 3 Cl2 ---> Al2Cl6 ---> mol Cl2 0.114 mol = 0.200 mol mol Al = 0.57 Limiting reactant = Cl 22 Limiting reactant = Cl Base all calcs. on Cl22 Base all calcs. on Cl calcs. grams Cl2 1 mol Al2Cl6 3 mol Cl2 This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio. Limiting reagent is grams Al2Cl6 Step 2: Calculate mass of Al 2Cl6 expected expected based on LR. 0.0380 mol Al 2Cl6 • 266.4 g Al2Cl6 = 1 0.1 g Al2Cl6 mol Cl2 Cl moles Cl2 moles Al2Cl6 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. was Therefore, Al was present in excess. But how much? • First find how much Al was required. First • Then find how much Al is in excess. is Calculating Excess Al Calculating Excess Al 2 Al + 3 Cl2 Cl 0.200 mol 0.200 mol products products Calculating Excess Al Calculating Excess Al 2 Al + 3 Cl2 Cl 0.200 mol 0.200 mol products products 0.114 mol = LR 0.114 mol = LR 0.114 mol = LR 0.114 mol = LR 2 mol Al = 0.0760 mol Al req'd 3 mol Cl2 0.114 mol Cl2 • Excess Al = Al available - Al required = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess Page 3 ...
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This note was uploaded on 01/11/2011 for the course ENGINEERIN MAE 107 taught by Professor Pozikrizdis during the Fall '08 term at San Diego.

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