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limiting reactants

# limiting reactants - Problem 1.64 of CAQS Construct the...

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Problem 1.64 of CAQS: Construct the reaction table for the reaction of 19.25 g V 2 O 5 ; 12.80 g C; and 30.66 g of Cl 2 to produce COCl 2 and VOCl 3 . In addition, calculate the mass of products and unreacted starting materials. Step 1 : Balance the reaction. 3 2 2 5 2 VOCl COCl Cl C O V + + + Work out for yourself that this turns out: 3 2 2 5 2 2 3 6 3 1 VOCl COCl Cl C O V + + + Step 2 : To work the reaction table, we need the initial amounts of reactants in mols. Thus, we require the molar masses viz: M(V2O5) = 181.8 g/mol M(C) = 12.0 g/mol M(Cl2) = 70.9 g/mol Thus, we get the following molar amounts: V2O5 = 0.106 mol C = 1.07 mol Cl2 = 0.432 mol So that our reaction table’s IN line would appear 1V2O5 3C 6Cl2 -> 3COCl2 2VOCl3 0.106 1.07 0.432 0 0 Step 3: Determine the limiting reagent. One may either proceed as does the book by calculating the amount of a product formed from each of the three reagents or, by using equivalents. I shall illustrate the latter here. Remember from lecture, that the # of equivalents = (# mols actually present / stoichiometric coefficient). Thus: #Eq(V2O5) = 0.106 #Eq(C) = 0.357 #Eq(Cl2) = 0.0720 (three sig. figures)

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