Percent efficiency.?

Question

You have equipment that operates at 61.0 percent efficiency for the reaction below.
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
If you need to produce 56.2 grams of iron in your equipment, how many grams of iron(III) oxide must you react with excess carbon monoxide?

I got 201g but it's not right. Any ideas. I really hate chemistry

maussy that wasn't right either

Answer 1

56.2 g of Fe correspond to 1 mole of iron. you see from the equation that 1 mole of Fe2O3 gives 2 moles of iron. So to have 1 mole of iron you must use 1/2 mole of Fe2O3
the mass of Fe2O3 is 2*56+3*16 =112+48=160g and 1/2 mole is 160/2 =80g of Fe2O3