Calculate the percent yield for the following reaction if 233g of NO2 reacted an

Question

Calculate the percent yield for the following reaction if 233g of NO2 reacted and 175g of HNO3 were produced .
3NO2 + H2O —-> 2HNO3 +NO
I have 46.01 g NO2
& 63.0g HNO3
How do I set this up
I have to include mol and show my work
actual yield theoretical yield X 100
I have to show how I got 82.2%
Would love to understand this better
Thank you Calculate the percent yield for the following reaction if 233g of NO2 reacted and 175g of HNO3 were produced .
3NO2 + H2O —-> 2HNO3 +NO
I have 46.01 g NO2
& 63.0g HNO3
How do I set this up
I have to include mol and show my work
actual yield theoretical yield X 100
I have to show how I got 82.2%
Would love to understand this better
Thank you
3NO2 + H2O —-> 2HNO3 +NO
I have 46.01 g NO2
& 63.0g HNO3
How do I set this up
I have to include mol and show my work
actual yield theoretical yield X 100
I have to show how I got 82.2%
Would love to understand this better
Thank you

Explanation / Answer

The reaction is--

3NO2 + H2O —-> 2HNO3 +NO

Now, let us calculate the theoretical yield of HNO3

(233 g NO2) x ( 1 mol NO2 / 46.0055 g NO2) x (2 mol HNO3/3 mol NO2) x ( 63.0130 g HNO3/1 mol HNO3) = 212.7 g HNO3  

Now,

percent yield =( actual yield/ theoretical yield ) x 100 % = (175 g / 212.7 g ) x 100 % = 0.822 x 100 % = 82.2 %

Required % yield = 82.2 %

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