Aluminum with a mass of 2.7 g has reacted with chloride acid. Calculate the volume of hydrogen released?

Let’s write the reaction equation:

2Al + 6HCl = 2AlCl3 + 3H2 ↑

Let’s find the amount of aluminum substance:

v (Al) = m (Al) / M (Al) = 2.7 / 27 = 0.1 (mol).

According to the reaction equation, 3 mol H2 is formed per 2 mol of Al, therefore:

v (H2) = v (Al) * 3/2 = 0.1 * 3/2 = 0.15 (mol).

Thus, the volume of released hydrogen, measured under normal conditions (n.o.):

V (H2) = v (H2) * Vm = 0.15 * 22.4 = 3.36 (l).

Answer: 3.36 l.