Question

Mining companies extract iron from iron ore according to the following balanced equation: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) In a reaction mixture containing 169 g Fe2O3 and 59.4 g CO, CO is the limiting reactant.

Part A Calculate the mass of the reactant in excess (which is Fe2O3) that remains after the reaction has gone to completion. Express the mass with the appropriate units.

Answer 1

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

no of moles of Fe2O3 = W/G.M.Wt

                                      = 169/159.69    = 1.0583 moles

no of moles of CO      = W/G.M.Wt

                                   = 59.4/28    = 2.12 moles

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

1 mole of Fe2O3 react with 3 moles of CO

1.0583 moles of Fe2O3 react with = 3*1.0583/1   = 3.1749 moles of CO is required

CO is limiting reactant

3 moles of CO react with 1 mole of Fe2O3

2.12 moles of CO react with = 1*2.12/3    = 0.7066 moles of Fe2O3 is required

Fe2O3 excess reacrtant

The no of moles of excess of Fe2O3 after complete the reaction = 1.0583-0.7066   = 0.3517 moles

The mass of exces of Fe2O3 after complete the reaction = no of moles * gram molar mass

                                                                                      = 0.3517*159.69   = 56.16g of Fe2O3 >>answer