When aluminum reacts with sulfuric acid, aluminum sulfate is formed and hydrogen is released: 2Al + 3H2SO4

When aluminum reacts with sulfuric acid, aluminum sulfate is formed and hydrogen is released: 2Al + 3H2SO4 → Al2 (SO4) 3 + 3H2. Calculate what mass of aluminum sulfate is formed by the interaction of aluminum with 2.94 g of sulfuric acid.

Given:
m (H2SO4) = 2.94 g
To find:
m (Al2 (SO4) 3
Decision:
3H2SO4 + 2Al = Al2 (SO4) 2 + 3H2
n (H2SO4) = m / M = 2.94 g / 98 g / mol = 0.03 mol
n (H2SO4): n (Al2 (SO4) 3) = 3: 1
n (Al2 (SO4) 3) = 0.03 mol / 3 = 0.01 mol
m (Al2 (SO4) 3) = n * M = 0.01 mol * 342 g / mol = 3.42 g
Answer: 3.42 g