MaxL221

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Jun 1, 2006
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What would be the percentage yield if 11.85 g of Na2S2O3 were obtained from a reaction mixture containing 12.60 g of Na2SO3 and 5.00 g of S?
Na2SO3 + S --> Na2S2O3




A. 55%
B. 60%
C. 65%
D. 70%
E. 75%

can some please explain this question step by step?
 

CJ122638

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Jun 7, 2006
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What is the answer? I think it should be E) 75%

So 12.6g of Na2SO3 is 0.1 mol
this reacts with 0.15625 mol of S (5/32)

this means that Na2SO3 is the limiting reagent.

So you expect to make 0.1 mol of Na2S2O3 which would be 15.8g

But you only made 11.85. 11.85/15.8 = 75%

Is that right?
 
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MaxL221

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Jun 1, 2006
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CJ122638 said:
What is the answer? I think it should be E) 75%

So 12.6g of Na2SO3 is 0.1 mol
this reacts with 0.15625 mol of S (5/32)

this means that Na2SO3 is the limiting reagent.

So you expect to make 0.1 mol of Na2S2O3 which would be 15.8g

But you only made 11.85. 11.85/15.8 = 75%

Is that right?
yep! thanks....great explination!