Aluminum metal reacts with sulfuric acid according to the following equation: 2Al(s)+ 3H₂SO₄(aq)→ Al₂(SO₄)₃(s)+ 3H₂(g)If 12.9 g of aluminum reacts with excess sulfuric acid, and 62.4 g of Al2(SO₄)₃ are collected, what is the percent yield of Al₂(SO₄)₃?

Question

Quizplus · Accepted Answer

First, we need to calculate the theoretical yield of Al2(SO4)3 using stoichiometry. The molar mass of Al is 26.98 g/mol, so 12.9 g Al is equal to 0.478 mol Al. According to the balanced equation, 2 mol Al produces 1 mol Al2(SO4)3, so 0.239 mol Al2(SO4)3 is produced. The molar mass of Al2(SO4)3 is 342.15 g/mol, so the theoretical yield is 81.5 g Al2(SO4)3.

The percent yield is calculated by dividing the actual yield (62.4 g Al2(SO4)3) by the theoretical yield (81.5 g Al2(SO4)3) and multiplying by 100%.

Percent yield = (62.4 g / 81.5 g) x 100% = 76.3%

Therefore, the best answer choice is C.

Aluminum Metal Reacts with Sulfuric Acid According to the Following