Hello, Olivia,
I don't have enough time to explain each step and there isn't enough room in this response box to add my table that describes each step. Sorry.
First, calculate the moles of both reactants. (I get 0.270 moles Al2O3 and 0.151 moles Fe). The limiting reactant is the Fe, since we need 9 moles of Fe per mole of 4 moles of Al2O3, and the ratio of what we actually have is below that amount. So all of the Fe will react before the Al2O3 is consumed. Using the moles of Fe, we need 4 times that amount of the Al2O3. The balanced equation says we get 8 moles of Al per 4 moles of Al2O3, so multiply the moles Al2O3 by that amount to get moles Al produced. Use the molar mass of Al to find grams. I get 7.23 grams of Al.
The equation says we'll get 3 moles of Fe3O4 per 9 moles of Fe, so multiply the moles of Fe by 9 to get moles Fe3O4. I get 0.050 moles Fe3O4, which amounts to 11.6 grams. If 10g was the actual amount produced, the percent yield is (10/11.6) or 86.2%.
I hope this helps a little. There are too many steps to outline many more details. The basic idea is to convert everything into moles and use the balanced equation to predict the moles of reactants consumed and products produced using the molar ratios in the equation. Then convert the moles into grams.
Bob
