Question

Calculate the maximum mass of iron that can be formed when 25.5 g of iron(III) oxide is react with excess carbon monoxide. (Ar: Fe=56, O=16, C=12)

Answer 1

Fe2O3 + 3CO = 2Fe + 3CO2

Mr(Fe2O3) = 2*Ar(Fe) + 3*Ar(O) = 2*56 +3*16 = 160 g/mol

n(Fe2O3) = m(Fe2O3)/Mr(Fe2O3) = (25.5 g)/(160 g/mol) = 0.159375 mol

n(Fe) = 2*n(Fe2O3) = 2*(0.159375 mol) = 0.31875 mol

m(Fe) = n(Fe)*Ar(Fe) = (0.31875 mol)*56(g/mol) = 17.85 g