Fe2O3 + 3CO -> 2Fe + 3CO2. When 240 g of Fe2O3 reacts with 84 g of CO, the maximum mass of Fe obtained is (Fe = 56, C = 12, O = 16)

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$Fe X_{2} O X_{3} + 3 CO 2 Fe + 3 CO X_{2}$ . When $240 g$ of $Fe X_{2} O X_{3}$ reacts with $84 g$ of CO, the maximum mass of Fe obtained is (Fe = 56, C = 12, O = 16)

A $168 g$
B $84 g$
C $112 g$
D $56 g$

Answer: C

n(Fe2O3) = 240/160 = 1.50 mol; n(CO) = 84/28 = 3.00 mol. CO needs 3 mol per mol Fe2O3, so Fe2O3 would need 4.50 mol CO. CO is limiting: Fe = (2/3) x 3 = 2 mol = 112 g.

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FeX2​OX3​+3CO​2Fe+3COX2​. When 240 g of FeX2​OX3​ reacts with 84 g of CO, the maximum mass of Fe obtained is (Fe = 56, C = 12, O = 16)

Answer: C

More questions on Limiting reagent, percent yield and percent excess

$Fe X_{2} O X_{3} + 3 CO 2 Fe + 3 CO X_{2}$ . When $240 g$ of $Fe X_{2} O X_{3}$ reacts with $84 g$ of CO, the maximum mass of Fe obtained is (Fe = 56, C = 12, O = 16)