200 g of 4% hydrochloric acid solution was added to 5 g of zinc. How much hydrogen will be released in this case?

1. The interaction of zinc with hydrochloric acid proceeds according to the equation:
Zn + 2HCl = ZnCl2 + H2 ↑;
2. find the mass of hydrogen chloride:
m (HCl) = w (HCl) * m (solution HCl) = 0.04 * 200 = 8 g;
3.Calculate the chemical quantities of the reacting substances:
n (HCl) = m (HCl): M (HCl) = 8: 36.5 = 0.2192 mol;
n (Zn) = m (Zn): M (Zn) = 5: 65 = 0.0769 mol;
4. zinc is taken in the deficiency, we determine the amount of released hydrogen:
n (H2) = n (Zn) = 0.0769 mol;
5. calculate the volume of hydrogen:
V (H2) = n (H2) * Vm = 0.0769 * 22.4 = 1.72 liters.
Answer: 1.72 liters.