Question

The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from the high-altitude jet plane, the SST. The reaction is O3 + NO longrightarrow O2 + NO2 If 0.781 g of O3 reacts with 0.589 g of NO, how many grams of NO2 will be produced?

Answer 1

Answer: 0.736 g

Step-by-step explanation:

`O_3+NO\rightarrow O_2+NO2`

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

`\text{Number of moles of}O_3=(0.781g)/(48g/mol)=0.016moles`

`\text{Number of moles of}NO=(0.589g)/(30g/mol)=0.019moles`

By Stoichiometry:

1 mole of ozone
`O_3` reacts with 1 mole of nitric oxide
`NO` to form 1 mole of nitrogen dioxide
`NO_2`

0.016 moles of ozone reacts with=
`(1)/(1)* 0.016=0.016moles` of nitric oxide to form 0.016 mole of
`NO_2`

Thus ozone is the limiting reagent as it limits the formation of products and nitric oxide is the excess reagent as (0.019-0.016) g= 0.003 g remains as such.

Mass of
`NO_2=moles* {\text{Molar mass}}=0.016* 46=0.736g`

0.736 g of
`NO_2` will be produced.