<u>Answer:</u>
<u>For a:</u> The number of molecules of nitrogen dioxide is ![4.52\times 10^{23}](https://tex.z-dn.net/?f=4.52%5Ctimes%2010%5E%7B23%7D)
<u>For b:</u> The mass of nitric acid formed is 54.81 grams
<u>For c:</u> The mass of nitric acid formed is 206 grams
<u>Explanation:</u>
The given chemical reaction follows:
![3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)](https://tex.z-dn.net/?f=3NO_2%28g%29%2BH_2O%28l%29%5Crightarrow%202HNO_3%28aq.%29%2BNO%28g%29)
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
of nitrogen dioxide
According to mole concept:
1 mole of a compound contains
number of molecules.
So, 0.75 moles of nitrogen dioxide will contain
number of molecules
Hence, the number of molecules of nitrogen dioxide is ![4.52\times 10^{23}](https://tex.z-dn.net/?f=4.52%5Ctimes%2010%5E%7B23%7D)
To calculate the number of moles, we use the equation:
.....(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}=\frac{60.0g}{46g/mol}=1.304mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20nitrogen%20dioxide%7D%3D%5Cfrac%7B60.0g%7D%7B46g%2Fmol%7D%3D1.304mol)
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 =
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\times 63g/mol)=54.81g](https://tex.z-dn.net/?f=0.870mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nitric%20acid%7D%7D%7B63g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20nitric%20acid%7D%3D%280.870mol%5Ctimes%2063g%2Fmol%29%3D54.81g)
Hence, the mass of nitric acid formed is 54.81 grams
- <u>For nitrogen dioxide:</u>
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}=\frac{225g}{46g/mol}=4.90mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20nitrogen%20dioxide%7D%3D%5Cfrac%7B225g%7D%7B46g%2Fmol%7D%3D4.90mol)
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}=\frac{55.2g}{18g/mol}=3.06mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20water%7D%3D%5Cfrac%7B55.2g%7D%7B18g%2Fmol%7D%3D3.06mol)
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 =
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
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\times 63g/mol)=206g](https://tex.z-dn.net/?f=3.27mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nitric%20acid%7D%7D%7B63g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20nitric%20acid%7D%3D%283.27mol%5Ctimes%2063g%2Fmol%29%3D206g)
Hence, the mass of nitric acid formed is 206 grams