How much sulfur needs to be taken in order for it to fully react with 2 g of aluminum?

How much sulfur needs to be taken in order for it to fully react with 2 g of aluminum? What mass of aluminum sulfide Al2S3 will be obtained?

1. Let’s write the reaction equation:

2Al + 3S = Al2S3.

2. Find the amount of aluminum:

n (Al) = m (Al) / M (Al) = 2 g / 27 g / mol = 0.074 mol.

3. Using the reaction equation, we find the amount of sulfur and sulfur sulfide, and then their mass:

3n (S) = 2n (Al)

n (S) = 2n (Al) / 3 = 2 * 0.074 mol / 3 = 0.05 mol.

m (S) = n (S) * M (S) = 0.05 mol * 32 g / mol = 1.6 g.

n (Al2S3) = 2n (Al) = 2 * 0.074 mol = 0.15 mol.

m (Al2S3) = n (Al2S3) * M (Al2S3) = 0.15 mol * 150 g / mol = 22.5 g.

Answer: m (S) = 1.6 g; m (Al2S3) = 22.5 g.