How much hydrogen will be released under the action of an excess of hydrochloric acid on 2.7 g

How much hydrogen will be released under the action of an excess of hydrochloric acid on 2.7 g of aluminum, if the hydrogen yield is 90% of the theoretical value?

Let’s implement the solution:

By the condition of the problem, we write down the equation of the process:
m = 2.7 g. X l. -? W = 90%

2Al + 6HCl = 2AlCl3 + 3H2 – OBP, hydrogen is evolved;

Calculations:
M (Al) = 26.9 g / mol.

M (H2) = 2 g / mol.

Y (Al) = m / M = 2.7 / 26.9 = 0.1 mol.

Proportion:
0.1 mol (Al) – X mol (H2);

-2 mol -3 mol from here, X mol (H2) = 0.1 * 3/2 = 0.15 mol.

We find the volume of H2:
V (H2) = 0.15 * 22.4 = 3.36 liters. (theoretical scope).

V (H2) = 0.90 * 3.36 = 3.02 l. (practical scope).

Answer: 3.02 liters of hydrogen was released.