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 ---> O2 + NO2
If 0.827 g of O3 reacts with 0.635 g of NO, how many grams of NO2 will be produced? g NO2 Which compound is the limiting reagent? ozone (O3) nitric oxide (NO) Calculate the number of moles of the excess reagent remaining at the end of the reaction.

Answer 1

Answer: Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

Explanation:

To calculate the number of moles, we use the equation

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$   ....(1)

• For ozone:

Given mass of ozone = 0.827 g

Molar mass of ozone = 48 g/mol

Putting values in above equation, we get:

$\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol$

• For nitric oxide:

Given mass of nitric oxide = 0.635 g

Molar mass of nitric oxide = 30.01 g/mol

Putting values in above equation, we get:

$\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol$

For the given chemical equation:

$O_3+NO\rightarrow O_2+NO_2$

By Stoichiometry of the reaction:

1 mole of ozone reacts with 1 mole of nitric oxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitric oxide

As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ozone is considered as a limiting reagent because it limits the formation of product.

• Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles

By Stoichiometry of the reaction:

1 mole of ozone produces 1 mole of nitrogen dioxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitrogen dioxide

Now, calculating the mass of nitrogen dioxide from equation 1, we get:

Molar mass of nitrogen dioxide = 46 g/mol

Moles of nitrogen dioxide = 0.0172 moles

Putting values in equation 1, we get:

$0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g$

Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.