Calculate the volume of hydrogen that is formed by the interaction of 4.8 g magnesium with 29.4 g sulfuric acid.

1. The interaction of magnesium with sulfuric acid proceeds according to the reaction equation:
Mg + H2SO4 = MgSO4 + H2 ↑;
2.Calculate the chemical amounts of magnesium and sulfuric acid:
n (Mg) = m (Mg): M (Mg) = 4.8: 24 = 0.2 mol;
n (H2SO4) = m (H2SO4): M (H2SO4);
M (H2SO4) = 2 + 32 + 4 * 16 = 98 g / mol;
n (H2SO4) = 29.4: 98 = 0.3 mol;
3. Determine the amount of released hydrogen by the lack, in our case by the amount of magnesium:
n (H2) = n (Mg) = 0.2 mol;
4.calculate the volume of hydrogen:
V (H2) = n (H2) * Vm = 0.2 * 22.4 = 4.48 liters.
Answer: 4.48 liters.