Zinc weighing 6.5 kg was treated with 120 kg of 10% sulfuric acid solution. Determine the amount of gas

Zinc weighing 6.5 kg was treated with 120 kg of 10% sulfuric acid solution. Determine the amount of gas evolved if the yield of this reaction product is 80% of the theoretically possible.

Given:
m (Zn) = 6.5 kg = 6500 g
m solution (H2SO4) = 120 kg = 120,000 g
ω (H2SO4) = 10%
η (gas) = 80%

Find:
V (gas) -?

Solution:
1) Zn + H2SO4 => ZnSO4 + H2 ↑;
2) n (Zn) = m (Zn) / M (Zn) = 6500/65 = 100 mol;
3) m (H2SO4) = ω (H2SO4) * m solution (H2SO4) / 100% = 10% * 120,000 / 100% = 12,000 g;
4) n (H2SO4) = m (H2SO4) / M (H2SO4) = 12000/98 = 122.45 mol;
5) n theory. (H2) = n (Zn) = 100 mol;
6) n practical (H2) = η (H2) * n theore. (H2) / 100% = 80% * 100/100% = 80 mol;
7) V (H2) = n (H2) * Vm = 80 * 22.4 = 1792 hp

Answer: The volume of H2 is 1792 liters.