Question

If you have 50 grams of H2SO4 and excess Al, how many grams of A₁₂(SO₄)₃ are produced during the following reaction?

2 Al + 3 H2SO4 - A₁₂(SO₄)₃+3 H2
A.58.2 g A₁₂(SO₄)₃
B.523.5 g A₁₂(SO₄)₃
C.174.5 g A₁₂(SO₄)₃
D.5700 g A₁₂(SO₄)₃

Answer 1

The mass of A₁₂(SO₄)₃ produced is 174.5 g.

Step-by-step explanation:

To find the mass of A₁₂(SO₄)₃ produced, we need to use the concept of stoichiometry. According to the balanced equation 2 Al + 3 H₂SO₄ → A₁₂(SO₄)₃ + 3 H₂, the molar ratio between Al and A₁₂(SO₄)₃ is 2:1. We start by converting the mass of H₂SO₄ to moles:

50 g H₂SO₄ * (1 mol / 98 g) = 0.51 mol H₂SO₄

Since Al is in excess, all of the 0.51 mol H₂SO₄ will react and form an equal moles of A₁₂(SO₄)₃. Finally, we can convert the moles of A₁₂(SO₄)₃ to grams:

0.51 mol A₁₂(SO₄)₃ * (342.2 g / mol) = 174.5 g A₁₂(SO₄)₃