Question

Iron(III) oxide reacts with carbon monoxide according to the equation:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
A reaction mixture initially contains 23.00 g Fe2O3and 15.66 g CO.

Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

Answer 1

Molar mass of Fe₂O₃ = (2x55.8) + (3x16) = 159.6 g/mol

Molar mass of CO = 12 + 16 = 28 g/mol

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

From the balanced equation ,

1 mole of Fe₂O₃ reacts with 3 moles of CO

                           OR

1x159.6 g of Fe₂O₃ reacts with 3x28 g of CO

23.00 g of Fe₂O₃ reacts with M g of CO

M = (23.00x3x28) / ( 1x159.6)

    = 12.10 g of CO

So 15.66 - 12.10 = 3.56 g of CO left unreacted so it is the excess reactant.