60 g of aluminum sulfide was treated with an excess of an aqueous solution of hydrochloric acid. Calculate the amount of gas released as a result of this reaction.
Al2S3 + 6HCl → 2AlCl3 + 3H2S ↑
We calculate the amount of aluminum sulfide substance:
n (Al2S3) = m (Al2S3) / M (Al2S3) = 60/150 = 0.4 mol,
Before Al2S3, the coefficient is 1, and H2S is 3. Therefore, the amount of H2S through the amount of Al2S3 can be expressed as follows:
n (H2S) = n (Al2S3) ⋅ 3/1 = 0.4 ⋅ 3/1 = 1.2 mol
V (H2S) = n (H2S) ⋅ Vm = 1.2 ⋅ 22.4 = 26.88 L ≈ 27 L
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