In the reaction of aluminum with sulfuric acid, 3.42 g of aluminum sulfate

In the reaction of aluminum with sulfuric acid, 3.42 g of aluminum sulfate was formed. Determine the mass and amount of sulfuric acid that has reacted

1. The formation of sodium sulfate occurs according to the reaction equation:

2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2 ↑;

2.Calculate the chemical amount of aluminum sulfate:

n (Al2 (SO4) 3) = m (Al2 (SO4) 3): M (Al2 (SO4) 3);

M (Al2 (SO4) 3) = 2 * 27 + 3 * 32 + 12 * 16 = 342 g / mol;

n (Al2 (SO4) 3) = 3.42: 342 = 0.01 mol;

3.determine the amount of sulfuric acid that has reacted:

n (H2SO4) = n (Al2 (SO4) 3) * 3 = 0.01 * 3 = 0.03 mol;

4. find the mass of the acid:

m (H2SO4) = n (H2SO4) * M (H2SO4);

M (H2SO4) = 2 + 32 + 4 * 16 = 98 g / mol;

m (H2SO4) = 0.03 * 98 = 2.94 g.

Answer: 2.94 g; 0.03 mol.