How much hydrogen will be released as a result of the interaction of 2.7 g aluminum with a sufficient

How much hydrogen will be released as a result of the interaction of 2.7 g aluminum with a sufficient amount of hydrochloric acid?

Let’s write down the solution, make up the equation:
2Al + 6HCl = 2AlCl3 + 3H2 – OBP, hydrogen gas is evolved;
Let’s determine the molecular weight of aluminum:
M (Al) = 26.9 g / mol.
We find the number of moles of aluminum, if its mass is known:
Y (Al) = m / M = 2.7 / 26.9 = 0.1 mol.
Let’s make the proportion:
0.1 mol (Al) – X mol (H2);
-2 mol -3 mol from here, X mol (H2) = 0.1 * 3/2 = 0.15 mol.
Let’s calculate the volume of hydrogen using Avogadro’s law:
1 mol of gas at normal level – 22.4 liters.
0.15 mol – X l (H2) hence, X l (H2) = 0.15 * 22.4 / 1 = 3.36 L.
Answer: The volume of hydrogen is 3.36 liters.