Question

When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain, and nitrogen oxide; 3NO2(g)+H20(l)-->2HNO3(aq)+NO(g). a) How many molecules of NO2 are needed to react with 0.250 mol of H2O? b) How many grams of HNO3 are produced when 60.0 g of NO2 completly reacts? c) How many grams of HNO3 can be produced if 225 g of NO2 is mixed with 55.2 g of H2O?

Answer 1

For a: The number of molecules of nitrogen dioxide is
`4.52* 10^(23)`

For b: The mass of nitric acid formed is 54.81 grams

For c: The mass of nitric acid formed is 206 grams

Step-by-step explanation:

The given chemical reaction follows:

`3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)`

• For a:

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with
`(3)/(1)* 0.250=0.75mol` of nitrogen dioxide

According to mole concept:

1 mole of a compound contains
`6.022* 10^(23)` number of molecules.

So, 0.75 moles of nitrogen dioxide will contain
`0.75* 6.022* 10^(23)=4.52* 10^(23)` number of molecules

Hence, the number of molecules of nitrogen dioxide is
`4.52* 10^(23)`

• For b:

To calculate the number of moles, we use the equation:

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

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

`\text{Moles of nitrogen dioxide}=(60.0g)/(46g/mol)=1.304mol`

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce =
`(2)/(3)* 1.304=0.870` moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

`0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol* 63g/mol)=54.81g`

Hence, the mass of nitric acid formed is 54.81 grams

• For c:

• For nitrogen dioxide:

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

`\text{Moles of nitrogen dioxide}=(225g)/(46g/mol)=4.90mol`

• For water:

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

`\text{Moles of water}=(55.2g)/(18g/mol)=3.06mol`

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with =
`(1)/(3)* 4.90=1.63mol` of water

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

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

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce
`(2)/(3)* 4.90=3.27mol` of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

`3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol* 63g/mol)=206g`

Hence, the mass of nitric acid formed is 206 grams