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
Al
2S
32mol 3mol 1mol
M(Al)=27g/mol
n(Al)=4,5g/27g/mol=0,17mol
x=0,17mol*1mol / 2 mol= 0,085mol
n(Al
2S
3)=0,085mol
M(Al
2S
3)=150g/mol
m(Al
2S
3)=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.