# In the reaction NO2(g) + O2(g) -> NO(g) + O3(g), 5.20 g of nitrogen dioxide and 3.90 of oxygen react.  Identify the limiting and excess reagents.

Finding the limiting reactant

Given

NO2(g) + O2(g) → NO(g) + O3(g)

5.20 g of nitrogen dioxide

3.90 g of oxygen.

First step balance the given chemical reaction

NO2(g) + O2(g) → 2NO(g) + O3(g)

This equation is combination of two different reaction so we cannot balance this reaction Please correct your reaction.

Lets consider the following reaction for example and i will let you know how to find the limiting reactant

2 NO2 + O2 → 2NO3

Second step will be finding the moles from the given mass.

Moles = mass  / molar mass

5.20 g of NO2  moles = 5.20/46 = 0.113

3.90 g of O2  moles = 3.90/32 = 0.121.

Third step divide the moles wiht the Stoichiometric coefficient from the balanced chemical reaction.

That is ...

for NO2 0.113/2 = 0.0565

for O2 0.121/1 = 0.121.

Fourth step compare the values and look for the smaller value.

NO2 has smaller value that is 0.0565 than O2 that is 0.121 so NO2 is the limiting reactant.

The amount of product formed will depend upon the amount of limiting reactant.

If we have to find the amount of the product we will use the moles of NO2 that is 0.113 and find the amount of NO3.

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In the reaction between nitrogen dioxide and oxygen, the products are nitrogen monoxide and ozone.

The chemical equation of the reaction is:

NO2(g) + O2(g) -> NO(g) + O3(g)

One mole of nitrogen dioxide reacts with one mole of oxygen. The reagents used in the reaction are 5.2 g of nitrogen dioxide and 3.9 g of oxygen. As the molar mass of nitrogen dioxide is 46, 5.2 g is equivalent to 0.113 moles. And as the molar mass of oxygen is 32, 3.9 g is equivalent to 0.121 moles.

As the number of moles of nitrogen dioxide is more than that of oxygen, the limiting reagent is nitrogen dioxide and the excess reagent is oxygen.

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