80 grams of iron (iii) oxide and 54 grams of aluminum react to form an elemental iron and aluminum oxide. in the lab you only manage to get 75% yield. a. how many grams of the non limiting reagent remains? b. how many grams of the limiting reagent remains? c. how many in grams of an elemental iron is formed?

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