Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.0 g of Fe2O3 react with excess Al ?

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First write down the balanced equation of reaction between the Fe2O3 and Al , one of the product is Fe so other product will be Fe2O3

Fe2O3 + 2Al  --- >  Al2O3 +  2Fe

Now, 28.56 g of Fe is the actual yield of Fe , in order to find the percent yield we will need theoretical yield of Fe too, that can be found using the numbers given is that 50 g of Fe2O3 is reacting with excess Al ,so Al is excess reactant and Fe2O3 is limiting reactant .So, the amount of Fe can be found using the balanced chemical equation :

1 mole of Fe2O3 forms 2 mol of Fe according to the balanced equation above, so how much Fe will be form when 50 g of Fe2O3 reacts? to find this first convert the grams of Fe2O3 to moles , simply divide the mass given with the molar mass of Fe2O3 ( 159.6 g/mol is the molar mass of Fe2O3).

50g x 1mol/159.6g = 0.313mol Fe2O3

.313 moles of Fe2O3 x 2 mole of Fe / 1 moles of Fe2O3 = 0.626 moles Fe

Now convert the moles of Fe to the grams :-

0.626moles of  Fe x 55.8g/mol = 34.93g Fe

finall find the percentage yield using the formula:

percentage yield = actaul yield / theoretical yield x100% = (28.65 / 34.93) x 100 = 82.02%