Aluminum metal reacts with sulfuric acid according to the following equation: 2Al(s) + 3H₂SO₄(aq) $\rightarrow$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

The first step is to determine the theoretical yield of Al2(SO4)3.

From the balanced equation: 
2 moles of Al produces 1 mole of Al2(SO4)3

Molar mass of Al: 26.98 g/mol 
Given mass of Al: 12.9 g 
Number of moles of Al: (12.9 g) / (26.98 g/mol) = 0.478 moles 
Theoretical yield of Al2(SO4)3: (0.478 moles Al) / 2 * (342.16 g/mol Al2(SO4)3) = 82.0 g

Now, we can calculate the percent yield:

Percent yield = (actual yield / theoretical yield) * 100%

Given mass of Al2(SO4)3: 62.4 g

Percent yield = (62.4 g / 82.0 g) * 100% = 76.3%

Therefore, the best choice is (C) 76.3%.

Aluminum Metal Reacts with Sulfuric Acid According to the Following →\rightarrow→

Aluminum Metal Reacts with Sulfuric Acid According to the Following $\rightarrow$