# Find the volume of gas released by the action of hydrochloric acid on 5.4 grams of aluminum.

Let’s find the amount of aluminum substance by the formula:

n = m: M.

M (Al) = 27 g / mol.

n = 5.4 g: 27 g / mol = 0.2 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

2Al + 6HCl = 2AlCl3 + 3H2 ↑.

According to the reaction equation, there is 3 mol of hydrogen for 2 mol of aluminum. Substances are in quantitative ratios 2: 3 = 1: 1.5.

The amount of hydrogen will be 1.5 times more than the amount of aluminum.

n (H2) = 1.5 n (Al) = 0.2 × 1.5 = 0.3 mol.

2 mol Al – 3 mol H2,

0.2 mol Al – х mol Н2,

x mol H2 = (0.2 mol × 3 mol): 2 mol = 0.3

Let’s find the volume of hydrogen.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 0.3 mol × 22.4 L / mol = 6.72 L.

Answer: 6.72 liters.