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Author: Subject: Stoichiometry question: Al₂S₃+6H₂O→2Al(OH)₃+3H₂S
hechz
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[*] posted on 7-10-2009 at 19:21
Stoichiometry question: Al₂S₃+6H₂O→2Al(OH)₃+3H₂S


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]
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[*] posted on 7-10-2009 at 21:10


Ugh, I tried for you but I got 12.96g :P

Sorry, I'll keep trying.




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[*] posted on 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.



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Alkyrisk
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[*] posted on 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!) ;)
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[*] posted on 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!
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Alkyrisk
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[*] posted on 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.:P
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