How many grams of Al2S3 can be obtained by the interactions of 2.7 g of Al and S

Let’s execute the solution:

According to the condition of the problem, we write the data:
m = 2.7 g X g -?

2Al + 3S = Al2S3 – compounds, obtained aluminum sulfide;

Calculations:
M (Al) = 26.9 g / mol;

M (Al2S3) = 149.8 g / mol;

Y (Al) = m / M = 2.7 / 26.9 = 0.1 mol.

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

-2 mol -1 mol from here, X mol (Al2S3) = 0.1 * 1/2 = 0.05 mol.

Find the mass of the product:
m (Al2S3) = Y * M = 0.05 * 149.8 = 7.49 g

Answer: the mass of aluminum sulfide is 7.49 g