How much Al is required for the reaction with hydrochloric acid to obtain 5.6L of H2?

Let’s implement the solution:
1. Let’s compose an equation according to the condition of the problem:
Y mole -? V = 5.6 l;
2Al + 6HCl = 2AlCl3 + 3H2 – OBP, hydrogen is evolved;
2. Let’s make the calculations:
M (Al) = 26.9 g / mol;
M (H2) = 2 g / mol.
3. Proportions:
1 mol of gas at normal level – 22.4 liters;
X mol (H2) – 5.6 liters. hence, X mol (H2) = 1 * 5.6 / 22.4 = 0.25 mol.
X mol (Al) – 0.25 mol (H2);
-2 mol – 3 mol from here, X mol (Al) = 2 * 0.25 / 3 = 0.16 mol.
Answer: to carry out the process, you need aluminum in the amount of 0.16 mol.