Aluminum weighing 5.4 g reacted with an excess of sulfur. The amount of substance of the reaction product will be …

1.Calculate the amount of aluminum substance by the formula:

n = m: M, where M is molar mass.

M (Al) = 27 g / mol.

n = 5.4 g: 27 g / mol = 0.2 mol.

2. Let’s compose the equation of the reaction between aluminum and sulfur, find the quantitative ratios of substances

2Al + 3S = Al2S3

When 2 mol of aluminum interacts with 3 mol of sulfur, 1 mol of aluminum sulfide Al2S3 is formed.

That is, for 2 mol of aluminum there is 1 mol of salt, which means that they react 2: 1, aluminum is 2 times more than aluminum sulfide. Therefore, the amount of aluminum sulfide substance will be 2 times less than that of aluminum, that is, 0.2: 2 = 0.1 mol.

3. Let’s make a proportion to find the amount of substance Al2S3.

2 mol (Al) – 1 mol (Al2S3),

0.2 mol (Al) – х mol (Al2S3),

x mol (Al2S3) = (0.2 mol × 1 mol): 2 mol = 0.1 mol.

Answer: n (Al2S3) = 0.1 mol.