In the interaction of 15 g of aluminum with sulfur, 35 g of aluminum sulfide

In the interaction of 15 g of aluminum with sulfur, 35 g of aluminum sulfide AL2 S3 was formed. How much is this percentage of the theoretically possible?

2Al + 3S = Al2S3 – compounds, aluminum sulfide is released;
Let’s make the calculations:
M (Al) = 26.9 g / mol;

M (Al2S3) = 149.8 g / mol;

Let’s determine the amount of metal:
Y (Al) = m / M = 15 / 26.9 = 0.56 mol.

Proportion:
0.56 mol (Al) – X mol (Al2S3);

-2 mol -1 mol from here, X mol (Al2S3) = 0.56 * 1/2 = 0.28 mol.

Find the mass and yield of the product:
m (Al2S3) = Y * M = 0.28 * 149.8 = 41.7 g (theoretical weight);

W = m (practical) / m (theoretical) * 100;

W = 35 / 41.7 * 100 = 83.93%

Answer: The aluminum sulfide salt is obtained, the product yield is 83.93%