The balanced equation is:Fe2O3+3CO→2Fe+3CO2
so 1 mole of iron (III) oxide requires 3 moles of CO in order to react completely. 8.65g is a lot less than 1 mole of Fe2O3.8.65gFe2O3∗(1molFe2O3/159.6gFe2O3)=0.054molFe2O3 so there's only 0.054moles of Fe2O3 available. Because of the ratio, you need three times as much CO, which is 0.162 moles of CO. Do you have enough CO?7.85gCO∗(1molCO2/8gCO)=0.280molCO There is a whole lot more CO than you need to react with all the Fe2O3, so the iron oxide is the LR, not the CO.The reactant that weighs less is not automatically the LR, it will be the reactant that runs out first. The LR is the reactant that gets you the least amount of product. Now, you need to use the iron oxide mass to calculate how much iron you SHOULD get, and compare that to how much the problem tells you gets made, and that's the percent yield.
