When aluminum interacts with a solution of sulfuric acid, the mass of the test tube decreased by 1.2 g

When aluminum interacts with a solution of sulfuric acid, the mass of the test tube decreased by 1.2 g, determine the mass of the formed salt

1. Let’s compose the equation of the proceeding reaction:

2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2 ↑;

2. The mass of the test tube decreases due to the release of hydrogen, we find its chemical amount:

n (H2) = m (H2): M (H2) = 1.2: 2 = 0.6 mol;

3. Determine the amount of formed aluminum sulfate:

n (Al2 (SO4) 3) = n (H2): 3 = 0.6: 3 = 0.2 mol;

4. Calculate the mass of the resulting salt:

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

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

m (Al2 (SO4) 3) = 0.2 * 342 = 68.4 g.

Answer: 68.4 g.