2.7 g of aluminum was treated with hydrochloric acid. How many liters of hydrogen will be released in this case

2.7 g of aluminum was treated with hydrochloric acid. How many liters of hydrogen will be released in this case, if the practical hydrogen yield is 90% of the theoretically possible.

Let’s execute the solution:

According to the condition of the problem, we write down the equation of the process:
2Al + 6HCl = 2AlCl3 + 3H2 – OBP, hydrogen is evolved;

We make 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.

Find the volume of the product:
V (H2) = 0.15 * 22.4 = 3.36 l (theoretical volume);

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

Answer: Taking into account the product yield, the volume of hydrogen was 3.02 liters.