a) 34 grams Al remains b) 20 grams Fe2O3 remains c) 42 grams Fe produced Iron(III) oxide is Fe2O3, so the balanced reaction is Fe2O3 + 2Al ==> Al2O3 + 2Fe Now, calculate the molar masses of the involved elements and compounds Atomic weight iron = 55.845 g/mol Atomic weight Aluminum = 26.981539 g/mol Atomic weight oxygen = 15.999 g/mol Molar mass Fe2O3 = 2 * 55.845 + 3 * 15.999 = 159.687 g/mol Now determine how many moles of each reactant you have. Moles Fe2O3 = 80 g / 159.687 g/mol = 0.500980042 mol Moles Al = 54 g / 26.981539 g/mol = 2.001368417 mol Since you need only 2 moles of Al per mole of Fe2O3, the limiting reactant is Fe2O3. So for every mole of Fe2O3 used, we get 2 moles of iron, so the mass of iron produced with 100% yield is double the moles of Fe2O3 multiplied by the atomic weight of iron, so: 0.500980042 mol * 2 * 55.845 g/mol = 55.95446089 g Additionally, for every mole of Fe2O3 consumed, 2 moles of Al are consumed. So the mass of consumed Al is 0.500980042 mol * 2 * 26.981539 g/mol = 27.03442508 g a. Since we only had 75% yield, only 75% of the calculated aluminum is consumed, so 54 g - 0.75 * 27.034425 = 33.72418125 Rounded to 2 significant figures gives 34g b. Since we only had 75% yield, we're left with 25% of the Fe2O3 which is 0.25 * 80 g = 20 g c. And we only get 75% of the iron, so 55.95446089 g * 0.75 = 41.96584567, rounded to 2 figures gives 42 grams