DAT Achiever - Test 2 #63 GC

[thejamespark]

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What would be the percentage yield if 11.85g of Na2S2O3 were obtained from a reaction mixture containing 12.60g of Na2So3 and 5g of S?
Na2SO3 + S ----> Na2S2O3

Answer is 75%

# of mole for Na2SO3 = 12.6 / 126g = 0.1mole
# mole for S = 5g / 32g = ~0.16mol [excess]

[ 11.85 / 158g x 0.1mole ] x 100 = 75%

I understand everything but why is it: (11.85 / 158g) x (1 / 0.1mole) and not just multiplying.... thats the confusing part for me. why are we dividing 0.1 mole.... any help would be appreciated.

 

[imbackasd]

Basically, you are seeing how many grams of product can you make with your limiting reagent (in this case Na2SO3). You get that by setting up a conversion from 0.1 mole to Grams of Na2SO3.

So 0.1 mole of Na2SO3 x 258 g/mole = 15.8 g of Na2SO3

Then they did 11.85 / 15.8 g.

They did the same thing but wrote it like (11.85/15.8 x 0.1) essentially the same thing.

 

[thejamespark]

dude ur a life saver. thanks so much. i knew that the Na2SO3 was the limiting but was stuck from there. REALLY helpful. thanks for clarifying 😉)