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Felny
Answers (2)
111g Fe * (1 mol Fe / 56g Fe ) * (1 mol Fe2O3 / 1 mol Fe) * (160g Fe2O3 / mol Fe2O3 ) = 159g Fe203
answer is B. 111g Fe * (1 mol Fe / 56g Fe ) * (1 mol Fe2O3 / 1 mol Fe) * (160g Fe2O3 / mol Fe2O3 )
