To solve this problem, we need to use the balanced chemical equation:
2Al + 3S -> Al2S3
Given that 1 mole of sulfur reacts, we can determine the number of moles of aluminum sulfide produced by using the stoichiometric ratio from the balanced equation. The ratio tells us that 2 moles of aluminum react with 3 moles of sulfur to produce 1 mole of aluminum sulfide.
So, if we have 1 mole of sulfur, we can calculate the number of moles of aluminum sulfide produced as follows:
(1 mole S) * (1 mole Al2S3 / 3 moles S) = 1/3 mol Al2S3
Therefore, if 1 mole of sulfur reacts, we will produce 1/3 mole of aluminum sulfide.
To calculate the mass of aluminum sulfide produced, we need to use the molar mass of Al2S3, which is the sum of the atomic masses of aluminum (Al) and sulfur (S).
The molar mass of Al2S3 is calculated as follows:
2(26.98 g/mol Al) + 3(32.06 g/mol S) = 2(26.98 g/mol) + 3(32.06 g/mol) = 54.02 g/mol Al2S3
Now we can calculate the mass of aluminum sulfide produced:
(1/3 mole Al2S3) * (54.02 g/mol Al2S3) = 18.0067 g Al2S3
Rounding to three significant figures, the mass of aluminum sulfide produced is approximately 18.0 g.
Therefore, the correct answer is 18.0 g, not 24.9 g as you mentioned.