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Ozone is an air pollutant produced when nitrogen dioxide, NO2, from automobile exhaust reacts with oxygen. The chemical equation for this reaction is:
NO2 + O2 = NO + O3
In a given reaction, 5.20g of nitrogen dioxide and 3.90g of oxygen react.
a) Identify the limiting and excess reagents in this case. Justify your prediction.
b) What mass of ozone is expected?
c) Only 4.42g of ozone was collected in the experiment. Calculate the percent yield in this reaction.
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NO2 + O2 --> NO + O3
In order to find the limiting reagent, first we must convert the mass in grams for each reagent into moles.
5.20 g NO2 (1 mole/46 g) = 0.113 moles NO2
3.90 g O2 (1 mole/32 g) = 0.122 moles O2
Since the two reactants react in a 1:1 molar manner, the reagent present with the smaller number of moles will be the limiting reagent. In this case, NO2 is present in the smaller number of moles, hence it will be the limiting reagent. That makes O2 the excess reagent.
The mass of ozone expected can be calculated from the number of moles of the limiting reagent (NO2). 0.113 moles of NO2 can produce 0.113 moles of O3, so multiply by the molecular weight to find of the theoretical mass of O3 produced:
0.113 moles O3 (48 g/1 mole) = 5.42 g O3
If only 4.42 g of ozone was actually collected from the reaction, we divide this amount by the theoretical yield and multiply by 100 to calculate the percent yield:
4.42 g/5.42 g * 100 = 81.5% yield.
So the answers are:
a) limiting reagent: NO2
excess reagent: O2
b) 5.42 g O3
c) 81.5% yield
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