Find the volume of gas that was released during the interaction of magnesium with 200 ml of a solution of dilute sulfuric

Find the volume of gas that was released during the interaction of magnesium with 200 ml of a solution of dilute sulfuric acid with a density of 1.4 and a mass fraction of 20%.

Given:
V solution (H2SO4) = 200 ml
ρ solution (H2SO4) = 1.4 g / ml
ω (H2SO4) = 20%

To find:
V (gas) -?

1) Mg + H2SO4 => MgSO4 + H2 ↑;
2) m solution (H2SO4) = ρ solution * V solution = 1.4 * 200 = 280 g;
3) m (H2SO4) = ω * m solution / 100% = 20% * 280/100% = 56 g;
4) M (H2SO4) = Mr (H2SO4) = Ar (H) * N (H) + Ar (S) * N (S) + Ar (O) * N (O) = 1 * 2 + 32 * 1 + 16 * 4 = 98 g / mol;
5) n (H2SO4) = m / M = 56/98 = 0.57 mol;
6) n (H2) = n (H2SO4) = 0.57 mol;
7) V (H2) = n * Vm = 0.57 * 22.4 = 12.8 liters.

Answer: The volume of H2 is 12.8 liters.