Fe can be prepared as 2 Al + Fe2O3 ----> 2 Fe + Al2O3. Suppose that 0.450 moles of Fe2O3 are reacted with an excess of Al. Suppose that 43.6 grams of Fe are obtained. What is the percent yield of Fe?

Question

a) 16.2%
b) 60.7%
c) 86.7%
d) 43.4%

Percent yield= actual yield of product divided by theoretical yield of product times 100.

The actual yield is stated to be 43.6 g of Fe, so I just need to know how to get the theoretical yield from the information it already gives me from the problem.

Answer 1

theoritical yield: .45*(2/1) moles, so you get .90 moles Fe or .90*55 =about 50 grams in my head.

percent yield=44/50= about 88 percnet. Answer c.

Answer 2

86.7%