Using the equation, find the mass and amount of the substance sulfate, aluminum if 4.5 g of aluminum entered the reaction?

Let’s execute the solution:
1. According to the condition of the problem, write the data:
m = 4.5 g. X g. -? Y -?
2Al + 3S = Al2S3 – compounds, aluminum sulfide was obtained;
2. Let’s make the calculations:
M (Al) = 26.9 g / mol;
M (Al2O3) = 149.8 g / mol;
Y (Al) = m / M = 4.5 / 26.9 = 0.167 mol.
3. Proportion:
0.167 mol (Al) – X mol (Al2S3);
-2 mol -1 mol from here, X mol (Al2S3) = 0.167 * 1/2 = 0.08 mol.
4. Find the mass of the product:
m (Al2S3) = Y * M = 0.08 * 149.8 = 11.98 g.
Answer: Received aluminum sulfide weighing 11.98 g.