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

Explanation:

$O_3+NO\rightarrow O_2+NO2$

To calculate the moles, we use the equation:

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

$\text{Number of moles of}O_3=\frac{0.781g}{48g/mol}=0.016moles$

$\text{Number of moles of}NO=\frac{0.589g}{30g/mol}=0.019moles$

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=$\frac{1}{1}\times 0.016=0.016moles$ 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 $NO_2=moles\times {\text{Molar mass}}=0.016\times 46=0.736g$

0.736 g of $NO_2$ will be produced.