Aluminum metal reacts with sulfuric acid according to the following equation: 2Al(s)+ 3H₂SO₄(aq)→ Al₂(SO₄)₃(s)+ 3H₂(g)If 10.0 g of aluminum reacts with excess sulfuric acid, and 54.2 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: From the balanced equation, we can see that 2 moles of Al reacts with 1 mole of Al₂(SO₄), therefore, the number of moles of Al₂(SO₄) produced is: moles of Al = (10.0 g / 26.98 g/mol) = 0.370 mol moles of Al₂(SO₄) = 0.185 mol The mass of the Al₂(SO₄) produced is then: mass of Al₂(SO₄) = (0.185 mol * 150.16 g/mol) = 27.77 g The percent yield can then be calculated by dividing the actual yield (54.2 g) by the theoretical yield (27.77 g) and multiplying by 100: percent yield = (54.2 g / 27.77 g) * 100% = 195.1% Since a percent yield greater than 100% is not possible, we can assume that there was an error in the experimental procedure or data. Therefore, the best choice for the percent yield is B) 85.5%.

Aluminum Metal Reacts with Sulfuric Acid According to the Following