Fe2O3(s)+3CO(g)=======>2Fe(s)+3CO2(g) 50.0g Fe2O3*(1 mol Fe2O3/ 159.69 g Fe2O3)*(2 mol Fe/1 mol Fe2O3)*(55.85 g Fe/1 mol Fe) 34.97 g Fe [ theoretical yield] % of yield = practical/theoretical * 100 = 32.8/34.97 *100 = 93.80 % Solution Fe2O3(s)+3CO(g)=======>2Fe(s)+3CO2(g) 50.0g Fe2O3*(1 mol Fe2O3/ 159.69 g Fe2O3)*(2 mol Fe/1 mol Fe2O3)*(55.85 g Fe/1 mol Fe) 34.97 g Fe [ theoretical yield] % of yield = practical/theoretical * 100 = 32.8/34.97 *100 = 93.80 %.