<u>Answer:</u> The percent yield of the nitrogen gas is 11.53 %.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of NO = 11.5 g
Molar mass of NO = 30 g/mol
Putting values in equation 1, we get:
![\text{Moles of NO}=\frac{11.5g}{30g/mol}=0.383mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20NO%7D%3D%5Cfrac%7B11.5g%7D%7B30g%2Fmol%7D%3D0.383mol)
- <u>For
:</u>
Given mass of
= 102.1 g
Molar mass of
= 92 g/mol
Putting values in equation 1, we get:
![\text{Moles of }N_2O_4=\frac{102.1g}{92g/mol}=1.11mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DN_2O_4%3D%5Cfrac%7B102.1g%7D%7B92g%2Fmol%7D%3D1.11mol)
For the given chemical reactions:
......(2)
.......(3)
- <u>Calculating the experimental yield of nitrogen gas:</u>
By Stoichiometry of the reaction 3:
6 moles of NO is produced from 2 moles of ![N_2O_4](https://tex.z-dn.net/?f=N_2O_4)
So, 0.383 moles of NO will be produced from =
of ![N_2O_4](https://tex.z-dn.net/?f=N_2O_4)
By Stoichiometry of the reaction 2:
1 mole of
produces 3 moles of nitrogen gas
So, 0.128 moles of
will produce =
of nitrogen gas
Now, calculating the experimental yield of nitrogen gas by using equation 1, we get:
Moles of nitrogen gas = 0.384 moles
Molar mass of nitrogen gas = 28 g/mol
Putting values in equation 1, we get:
![0.384mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.384mol\times 28g/mol)=10.75g](https://tex.z-dn.net/?f=0.384mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nitrogen%20gas%7D%7D%7B28g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20nitrogen%20gas%7D%3D%280.384mol%5Ctimes%2028g%2Fmol%29%3D10.75g)
- <u>Calculating the theoretical yield of nitrogen gas:</u>
By Stoichiometry of the reaction 2:
1 mole of
produces 3 moles of nitrogen gas
So, 1.11 moles of
will produce =
of nitrogen gas
Now, calculating the theoretical yield of nitrogen gas by using equation 1, we get:
Moles of nitrogen gas = 3.33 moles
Molar mass of nitrogen gas = 28 g/mol
Putting values in equation 1, we get:
![3.33mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(3.33mol\times 28g/mol)=93.24g](https://tex.z-dn.net/?f=3.33mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nitrogen%20gas%7D%7D%7B28g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20nitrogen%20gas%7D%3D%283.33mol%5Ctimes%2028g%2Fmol%29%3D93.24g)
- To calculate the percentage yield of nitrogen gas, we use the equation:
![\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20yield%7D%3D%5Cfrac%7B%5Ctext%7BExperimental%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%5Ctimes%20100)
Experimental yield of nitrogen gas = 10.75 g
Theoretical yield of nitrogen gas = 93.24 g
Putting values in above equation, we get:
![\%\text{ yield of nitrogen gas}=\frac{10.75g}{93.24g}\times 100\\\\\% \text{yield of nitrogen gas}=11.53\%](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20yield%20of%20nitrogen%20gas%7D%3D%5Cfrac%7B10.75g%7D%7B93.24g%7D%5Ctimes%20100%5C%5C%5C%5C%5C%25%20%5Ctext%7Byield%20of%20nitrogen%20gas%7D%3D11.53%5C%25)
Hence, the percent yield of the nitrogen gas is 11.53 %.