60 g of aluminum sulfide was treated with an excess of an aqueous solution of hydrochloric acid.

60 g of aluminum sulfide was treated with an excess of an aqueous solution of hydrochloric acid. calculate the volume of gas released as a result of this reaction.

Let’s write the reaction equation:

Al2S3 + 6HCl = 2AlCl3 + 3H2S ↑

Let’s find the amount of aluminum sulfide substance:

v (Al2S3) = m (Al2S3) / M (Al2S3) = 60/150 = 0.4 (mol).

According to the reaction equation, for 1 mol of Al2S3, 3 mol of H2S are formed, therefore:

v (H2S) = v (Al2S3) * 3 = 0.4 * 3 = 1.2 (mol).

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

V (H2S) = v (H2S) * Vm = 1.2 * 22.4 = 26.88 (l).