hechz - 7-10-2009 at 19:21
Al₂S₃+6H₂O→2Al(OH)₃+3H₂S
Concerning the above reaction I am asked what mass of Al(OH)₃ will be produced when 14.2g of Al₂S₃is fully reacted.
My first step was to convert the mass to the count of moles.
14.2g
----------------- = 0.115 mol of Al₂S₃
123.177g/mol
Then I determine the number of moles of the product, using the coefficients of the balanced equation:
⎛ 2 mol/Al(OH)₃ ⎞
0.115 mol/Al₂S₃⎜ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ⎟ = 0.23
mol/Al(OH)₃
⎝ 1
mol/AlS₃ ⎠
0.23(78.006)=17.94g/Al(OH)₃
The textbook answer key states the answer as 14.7g/Al(OH)₃
I am failing to see wherein my error lies.
[Edited on 8-10-2009 by hechz]
[Edited on 8-10-2009 by hechz]
psychokinetic - 7-10-2009 at 21:10
Ugh, I tried for you but I got 12.96g 
Sorry, I'll keep trying.
woelen - 7-10-2009 at 22:52
14.2 grams of Al2S3 is not 0.115 mol, but 0.0946 mol. With this info you should be able to find your mistake.
Alkyrisk - 15-11-2009 at 04:34
Molar mass for Al2S3 is 150.158 g/mol. Why? because
26.982 x 2 = 53.964
32.065 x 3 = 96.195
150.159 - 123.177(your hypothetical molar mass) =
????...??? exactly 26.982 (What these numbers mean? This is the question!) 
Nicodem - 15-11-2009 at 05:24
I don't know what kind of acrobatic arithmetics you are doing above and I'm not going to search for where you made a mistake but I think Woelen
already pointed it out. Instead of using numbers which leads to confusion, use variables instead. Like this way:
m[Al(OH)3] = M[Al(OH)3] * 2 * m[Al2S3] / M[Al2S3]
When m[Al2S3] = 14.2 g then m[Al(OH)3] = 14.75 g just like your textbook claims!
Alkyrisk - 15-11-2009 at 07:03
I want to find the origin of mistake. He took only one aluminium molecule instead 2.
M Al2S3 = 150.158, M (AlS3) = 123.177. The confusion may be derived from my acrobatic poor english.