Calculate the mass of Al sulfide (Al2S3) if 4.5 g (Al) entered the reaction.

1.2Al + 3S = Al2S3;

2.n (Al) = m (Al): M (Al) = 4.5: 27 = 0.167 mol;

3.the amounts of aluminum and its sulfide are in the ratio 2: 1, therefore:

n (Al2S3) = n (Al): 2 = 0.167: 2 = 0.0835 mol;

4.find the mass of aluminum sulfide:

m (Al2S3) = n (Al2S3) * M (Al2S3);

M (Al2S3) = 2 * 27 + 3 * 32 = 150 g / mol;

m (Al2S3) = 0.0835 * 150 = 12.525 g.

Answer: 12.525 g.