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only 4.42 g of ozone was collected in the experiment. Calculate the percent yield in this reaction. Ozone is an air pollutant produced when nitrogen dioxide, NO2 , from automobile exhaust reactswith oxygen. The chemical equation for this reaction is NO2(g) + o2(g) ---> NO(g) + O3(g). in a given reaction, 5.20g of nitrogen dioxide and 3.90g of oxygen react.

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Percent yield is a way to determine the efficiency of the reaction.

Theoretical yield (TY): computed value

Actual yield(AY): value obtained from the lab.

AY will NEVER be greater than TY at any circumstances.

to compute for the percent yield:


Percent yield = (mass Actual Yield/ mass theoretical yield)  x 100

In this problem AY is given with the value 4.42 grams of ozone.We need to look for the theoretical yield of ozone. In order to do that we have to employ stoichiometry. 

First step: Check if the reaction equation is balanced. It is balanced.

Second Step: Identify the limiting reagent. (this is the reagent that will produce the least number of O3 moles when calculated)


5.20 g NO2 x (1mole NO2/ 46gNO2) x (1mole O3/1mole NO2)

= 0.11304 moles O3 ----> limiting reagent

3.9 g O2 x (1mole O2/ 32g O2) x (1mole O3/ 1 mole O2)

=0.12188 moles O3

From the value of O3 derived from the limiting reagent, we get the mass of O3

 0.11304 moles O3 x (48 g O3/ 1mole O3)

= 5.4261g O3 ----------> mass theoretical yield


Now we are ready to compute the percent yield :)

Percent yield = (mass Actual Yield/ mass theoretical yield)  x 100

Percent yield = (4.42/5.4261)  x 100

                          81.5%  ---> final answer


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