[auksas]

4.5 g of aluminum is mixed with sulfur. How many grams of aluminum sulfide were obtained?

0,17mol
4,5g                        x mol, x g
2Al+3S Al₂S₃
2mol 3mol          1mol

M(Al)=27g/mol
n(Al)=4,5g/27g/mol=0,17mol

x=0,17mol*1mol / 2 mol= 0,085mol
n(Al₂S₃)=0,085mol
M(Al₂S₃)=150g/mol
m(Al₂S₃)=0,085mol*150g/mol=12,75g

The answer of this task is 24,9 g Al2S3. Where am I wrong? I can solve other tasks using the same method, but this time I've got the wrong answer.

[Hunter2]

Your result is correct. The given answer in the task is 2 times to large.

[auksas]

Your result is correct. The given answer in the task is 2 times to large.

But why? When such steps to right solution must be done?

[Hunter2]

I would say is a mistyping or mis calculation.

[Borek]

Must be a mistake in the answer key.

Note though: don't round down intermediate results, only the final answer. Correct result is not 12.75g but 12.52g.

Technically it should be rounded down to 2 significant digits, so 13g in both cases. Some teachers treat significant digits almost religiously. In reality they don't matter much, kind of a poor man's way of expressing the accuracy. There are better and much more strict ways.

[auksas]

I would say is a mistyping or mis calculation.

Obviously. But I can't find where

[auksas]

Must be a mistake in the answer key.

Note though: don't round down intermediate results, only the final answer. Correct result is not 12.75g but 12.52g.

Technically it should be rounded down to 2 significant digits, so 13g in both cases. Some teachers treat significant digits almost religiously. In reality they don't matter much, kind of a poor man's way of expressing the accuracy. There are better and much more strict ways.

Thanks

[Barbara Lewis]

Yes, I also think that you gave the right variant and the give one was a typo.
Given:
4.5 g Al
Balanced equation:
2Al + 3S → Al2S3
Moles of Al:
4.5 g Al / 27 g/mol Al = 0.17 mol Al
Using stoichiometry:
0.17 mol Al * (1 mol Al2S3 / 2 mol Al) = 0.085 mol Al2S3
Grams of Al2S3:
0.085 mol Al2S3 * 150 g/mol Al2S3 = 12.75 g Al2S3
Therefore, unless there is additional information missing, the final mass of Al2S3 obtained from reacting 4.5 g Al with excess S should be 12.75 g.