GCHEM Question

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jdoc04

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Kaplan Full Length Exam 1

MnO4 + I + H --> Mn + H20 + I2
Balanced as:
2 MnO4 + 10 I + 16 H --> 2 Mn + 5 I2 + 8 H20
If 150mg of KI is reacted & 108.5I2 is produced, what is the percent yield for the rxn?

Kaplan explanation:
It is evident that 5 moles of I2 will be produced, theoretically, for every 10 moles of I reacted. The stoichiometric calculation of the theoretical yield is then:
127mgI/166mgKI *150mgKI = 115mgI
%Y=actual/theoretical --> 108.5/115=0.945 --> 94.5%

I know the & yield formula, how to balance the eq., & how to go about using the stoichiometry to find the theoretical value. I am thrown off by the units (mg) vs (g). I weights 126.9 g/mol K weights 39.1 g/mol They just took the gram weight and made them equal to the mg weights.

My guess is it does not matter whether the problem is listed in mg or g. I guessed and just moved b/c the conversion made for a more difficult calculation.

Anyone??
 
Kaplan Full Length Exam 1

MnO4 + I + H --> Mn + H20 + I2
Balanced as:
2 MnO4 + 10 I + 16 H --> 2 Mn + 5 I2 + 8 H20
If 150mg of KI is reacted & 108.5I2 is produced, what is the percent yield for the rxn?

Kaplan explanation:
It is evident that 5 moles of I2 will be produced, theoretically, for every 10 moles of I reacted. The stoichiometric calculation of the theoretical yield is then:
127mgI/166mgKI *150mgKI = 115mgI
%Y=actual/theoretical --> 108.5/115=0.945 --> 94.5%

I know the & yield formula, how to balance the eq., & how to go about using the stoichiometry to find the theoretical value. I am thrown off by the units (mg) vs (g). I weights 126.9 g/mol K weights 39.1 g/mol They just took the gram weight and made them equal to the mg weights.

My guess is it does not matter whether the problem is listed in mg or g. I guessed and just moved b/c the conversion made for a more difficult calculation.

Anyone??

As long as your units are consistent it makes no difference whether you are using g, mg, microg or nanograms.
 
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