Fe2o3(s)+3co(g)→2fe(s)+3co2(g) in a reaction mixture containing 189 g fe2o3 and 63.0 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.

69.3 g First, determine the molar masses involved: Atomic weight iron = 55.845 Atomic weight carbon = 12.0107 Atomic weight oxygen = 15.999 Molar mass Fe2O3 = 2 * 55.845 + 3 * 15.999 = 159.687 g/mol Molar mass CO = 12.0107 + 15.999 = 28.0097 g/mol Determine how many moles of each reactant we have Moles Fe2O3 = 189 g / 159.687 g/mol = 1.18356535 mol Moles CO = 63.0 g / 28.0097 g/mol = 2.249220806 mol For every mole of Fe2O3, we need 3 moles of CO. So let's see how many moles of Fe2O3 is consumed by dividing moles CO by 3. 2.249220806 mol / 3 = 0.749740269 mol So we'll be consuming 0.749740269 moles of Fe2O3, subtract that from what we started with 1.18356535 mol - 0.749740269 mol = 0.433825081 mol Now multiply by the molar mass of Fe2O3 0.433825081 mol * 159.687 g/mol = 69.27622574 g Rounding to 3 significant figures gives 69.3 g