posted by CJ .
In the reaction below, 16 g of H2S with excess O2 produced 8 g of sulfur.
? H2S + ? O2 → ? S + ? H2O .
What is the percent yield of sulfur?
Answer in units of %
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2H2S + O2 ==> S + 2H2O
mols H2S = grams/molar mass
Use the coefficients in the balanced equation to convert mols H2S to mols S. (hint: so mols S = 1/2 mols H2S.)
Convert mols S to grams S. grams = mols x molar mass. This is the theoretical yield (TY). The actual yield(AY) is 8.0g from the problem.
%yield = (AY/TY)*100 = ?
I still don't understand how to get the theroetical yield.
I gave you a step by step procedure.
mols H2S = grams H2S/molar mass H2S = 16/34 = about 0.47 mols H2S.
Now convert mols H2S to mols S. In the balanced equation you have 2 mols H2S forming 2 mols S (yes, I made a typo. The balanced equation is
2H2S + O2 ==> 2S + 2H2O so 0.47 mols H2S = 0.47 mols S.
Convert to grams S or g = mols x atomic mass S = 0.47 x 32 = about 15 and this is the theoretical yield which is about twice if you followed the wrong equation I posted first.
%yield = (8/15) = about 53%
You can go through and clean up my estimates here and there.