Ammonium perchlorate, NH4ClO4, is used with aluminum as rocket fuel. $10 \mathrm{Al}(s)+6 \mathrm{NH}_{4} \mathrm{ClO}_{4} \rightarrow 4 \mathrm{Al}_{2} \mathrm{O}_{3}+2 \mathrm{AlCl}_{3}+12 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{~N}_{2}$ If 100.0 g of aluminum are reacted with 60.0 g of ammonium perchlorate, how many grams of aluminum oxide are produced?

Question

Quizplus · Accepted Answer

The balanced chemical equation shows that 10 moles of Al react with 6 moles of NH4ClO4 to produce 4 moles of Al2O3. Therefore, the mole ratio of Al to Al2O3 is 10:4, or 5:2. This means that for every 5 moles of Al reacted, 2 moles of Al2O3 are produced.To convert 100.0 g of Al to moles, we divide by the molar mass of Al (26.98 g/mol):100.0 g Al / 26.98 g/mol = 3.708 mol AlTo convert 60.0 g of NH4ClO4 to moles, we divide by the molar mass of NH4ClO4 (117.49 g/mol):60.0 g NH4ClO4 / 117.49 g/mol = 0.511 mol NH4ClO4Comparing the moles of Al and NH4ClO4, we see that Al is in excess. Therefore, the limiting reactant is NH4ClO4.Using the mole ratio of Al to Al2O3 (5:2), we can calculate the moles of Al2O3 produced:0.511 mol NH4ClO4 * (2 mol Al2O3 / 6 mol NH4ClO4) = 0.170 mol Al2O3Finally, we convert the moles of Al2O3 to grams by multiplying by the molar mass of Al2O3 (101.96 g/mol):0.170 mol Al2O3 * 101.96 g/mol = 17.35 g Al2O3Therefore, the correct answer is A) 34.71 g.

Ammonium Perchlorate, NH4ClO4, Is Used with Aluminum as Rocket Fuel