When aluminum interacts with sulfur, 45 g of aluminum sulfide was deposited. How many grams of sulfur have reacted?

Let’s write down the solution by composing the reaction equation:
2Al + 3S = Al2S3 – compound reaction, aluminum sulfide was obtained;
Let’s make calculations using the formulas of substances:
M (Al) = 26.9 g / mol;
M (Al2S3) = 2 * 26.9 + 3 * 32 = 149.8 g / mol;
Let’s calculate the number of moles of aluminum:
Y (Al) = m / M = 4.5 / 149.8 = 0.3 mol;
Let’s make the proportion:
X mol (Al) – 0.3 mol (Al2S3);
-2 mol -1 mol hence, X mol (Al) = 2 * 0.3 / 1 = 0.6 mol;
We find the mass of aluminum by the formula:
m (Al) = Y * M = 0.6 * 26.9 = 16.14 g.
Answer: the mass of aluminum that has reacted is 16.14 g.