Just checking an answer...
I was given Aluminum + Sulfur yields.. & I was to finish & balance & answer the questions.
2Al + 3S => Al2S3
If 1 mole of sulur reacts, how many moles of aluminum sulife are produced?
.166 mol Al2S3
How many grams of aluminum sulfide is this?
24.9 g
Has your teacher talked about sulfur occurring naturally as S8?
That would change what you have done. If s/he hasn't talked about that, you still don't have mols Al2S3 correct. Based on your equation I think 1 mol S x (1 mol Al2S3/3 mol S) = 0.333 mol Al2S3 is the correct answer. Grams of Al2S3 are correct based on your answer of 0.166 mol Al2S3.so your method must be ok for that part of the problem.
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.