If 15.0g of aluminum is reacted with 30.0g of iron(III) oxide according to the f
ID: 998795 • Letter: I
Question
If 15.0g of aluminum is reacted with 30.0g of iron(III) oxide according to the following unbalanced equation Al+Fe2O3=Fe+Al2O3 What is the limiting reactant? What is the theoretical yield (in grams) of iron metal? What is the percent yield if 19.8 is the experimental yield? If 15.0g of aluminum is reacted with 30.0g of iron(III) oxide according to the following unbalanced equation Al+Fe2O3=Fe+Al2O3 What is the limiting reactant? What is the theoretical yield (in grams) of iron metal? What is the percent yield if 19.8 is the experimental yield? What is the limiting reactant? What is the theoretical yield (in grams) of iron metal? What is the percent yield if 19.8 is the experimental yield?Explanation / Answer
2Al + Fe2O3-------->2Fe+Al2O3
2 moles 1 mole
no of moles of Al = W/G.A.Wt
= 15/27 = 0.56 moles
no of moles of Fe2O3 = W/G.M.Wt
= 30/159.7 = 0.187 moles
2 mole of Al react with 1moles of Fe2o3
0.56 moles of Al react with = 1*0.56/2 = 0.28 moles of Fe2O3 is required
Fe2O3 is limiting reagent or limiting reactant
1 mole of Fe2O3 react with Al to form 2 moles of Fe
159.7 gm of Fe2O3 react with Al to form 2* 56gm of Fe
30gm of Fe2O3 react with Al to form = 2*56*30/159.7 = 21.04 gm of Fe
Theoretical yield of iron = 21.04gm
percentage yield = actual yield*100/theoritical yield
= 19.8*100/21.04 = 94.1%
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