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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?

Respuesta :

Answer:

For a: The number of molecules of nitrogen dioxide is [tex]4.52\times 10^{23}[/tex]

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

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

Explanation:

The given chemical reaction follows:

[tex]3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)[/tex]

  • 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 [tex]\frac{3}{1}\times 0.250=0.75mol[/tex] of nitrogen dioxide

According to mole concept:

1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules.

So, 0.75 moles of nitrogen dioxide will contain [tex]0.75\times 6.022\times 10^{23}=4.52\times 10^{23}[/tex] number of molecules

Hence, the number of molecules of nitrogen dioxide is [tex]4.52\times 10^{23}[/tex]

  • For b:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol[/tex]

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 = [tex]\frac{2}{3}\times 1.304=0.870[/tex] 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:

[tex]0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g[/tex]

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:

[tex]\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol[/tex]

  • For water:

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol[/tex]

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 = [tex]\frac{1}{3}\times 4.90=1.63mol[/tex] 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 [tex]\frac{2}{3}\times 4.90=3.27mol[/tex] 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:

[tex]3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g[/tex]

Hence, the mass of nitric acid formed is 206 grams

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