Finding the limiting reactant
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.