For the following reaction:

Fe2O3 + 3CO --> 2Fe + 3CO2
how many grams of Fe2O3 is needed to produce 111 g of Fe?

A) 79.3 g Fe2O3
B) 159 g Fe2O3
C) 317 g Fe2O3
D) 71.4 g Fe2O3

3 answers

? g = 111 g Fe * 1 mol Fe2O3/2 mol Fe * 159.6 g/1 mol Fe2O3
= 8857.8 g of Fe2O3
answer is B.

111g Fe * (1 mol Fe / 56g Fe ) * (1 mol Fe2O3 / 1 mol Fe) * (160g Fe2O3 / mol Fe2O3 )
111g Fe * (1 mol Fe / 56g Fe ) * (1 mol Fe2O3 / 1 mol Fe) * (160g Fe2O3 / mol Fe2O3 )

= 159g Fe203