Question

Ozone, O3, is produced in automobile exhaust by the reaction represented by the equation:

NO2 (g) + O2 (g) → NO (g) + O3 (g)
What mass of ozone is predicted to form from the reaction of 5.0 g NO2 in a car’s exhaust and excess oxygen?

Answer 1

Answer: 5.28 g of
`O_3` is produced.

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} NO_2=(5.0g)/(46g/mol)=0.11moles`

The balanced chemical reaction is:

`NO_2(g)+O_2(g)\rightarrow NO(g)+O_3(g)`

`NO_2` is the limiting reagent as it limits the formation of product and
`O_2` is the excess reagent.

According to stoichiometry :

1 mole of
`NO_2` produces = 1 mole of
`O_3`

Thus 0.11 moles of
`NO_2` will produce=
`(1)/(1)* 0.11=0.11moles` of
`O_3`

Mass of
`O_3=moles* {\text {Molar mass}}=0.11moles* 48g/mol=5.28g`

Thus 5.28 g of
`O_3` is produced.