What volume of hydrogen is formed by the interaction of hydrochloric acid with 50 g of aluminum?

Given:
m (Al) = 50 g
Vm = 22.4 l / mol

Find:
V (H2) -?

Solution:
1) Write the equation of a chemical reaction:
6HCl + 2Al => 2AlCl3 + 3H2 ↑;
2) Calculate the molar mass of Al:
M (Al) = Mr (Al) = Ar (Al) = 27 g / mol;
3) Calculate the amount of Al substance:
n (Al) = m (Al) / M (Al) = 50/27 = 1.85 mol;
4) Determine the amount of substance H2:
n (H2) = n (Al) * 3/2 = 1.85 * 3/2 = 2.78 mol;
5) Calculate the volume of H2 (under normal conditions):
V (H2) = n (H2) * Vm = 2.78 * 22.4 = 62.3 liters.

Answer: The volume of H2 is 62.3 liters.