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

Answer 1

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 %