Question

If 8.65 grams of iron (III) oxide reacts with 7.85 grams of carbon monoxide to produce 5.25 g of pure iron, what are the theoretical yield and percent yield of this reaction? Be sure to show the work that you did to solve this problem.

unbalanced equation: Fe2O3 + CO Fe + CO2

Answer 1

I was able to balance the equation already...but if someone can help me with the equation for finding the percentage yields, that would be great;)

Answer 2

you're given amounts of both reactants, so this is a limiting reactant problem. Can you find which reactant is the limiting reactant?

Answer 3

.....Carbon monoxide?

Answer 4

The balanced equation is:\[Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\]
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.
\[8.65g Fe_2O_3 * (\frac{1mol Fe_2O_3}{159.6g Fe_2O_3}) = 0.054mol Fe_2O_3\] 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?
\[7.85g CO * (\frac{1mol CO}{28g CO}) = 0.280mol CO\] 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.

Answer 5

This helped, thank you!;)