What is the mass of aluminum (Al) in reaction with sulfur (S) if 0.1 mol of aluminum sulfide (Al2S3) is obtained?

Let’s write the reaction equation:

2Al + 3S (t) = Al2S3

By condition, the amount of aluminum sulfide substance:

v (Al2S3) = 0.1 (mol).

According to the reaction equation, 1 mol of Al2S3 is formed from 2 mol of Al, therefore:

v (Al) = v (Al2S3) * 2 = 0.1 * 2 = 0.2 (mol).

Thus, the mass of reacted aluminum is:

m (Al) = v (Al) * M (Al) = 0.2 * 27 = 5.4 (g).

Answer: 5.4 g.

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