Asked by Petey
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
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
Answers
Answered by
Anonymous
? g = 111 g Fe * 1 mol Fe2O3/2 mol Fe * 159.6 g/1 mol Fe2O3
= 8857.8 g of Fe2O3
= 8857.8 g of Fe2O3
Answered by
Felny
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 )
Answered by
Felny
111g Fe * (1 mol Fe / 56g Fe ) * (1 mol Fe2O3 / 1 mol Fe) * (160g Fe2O3 / mol Fe2O3 )
= 159g Fe203
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