Question

During studies of the reaction below,

Answer 1

Answer: The percent yield of the nitrogen gas is 11.53 %.

Explanation:

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

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

• For NO:

Given mass of NO = 11.5 g

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

$\text{Moles of NO}=\frac{11.5g}{30g/mol}=0.383mol$

• For $N_2O_4$ :

Given mass of $N_2O_4$ = 102.1 g

Molar mass of $N_2O_4$ = 92 g/mol

Putting values in equation 1, we get:

$\text{Moles of }N_2O_4=\frac{102.1g}{92g/mol}=1.11mol$

For the given chemical reactions:

$2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)$      ......(2)

$N_2H_4(l)+2N_2O_4(l)\rightarrow 6NO(g)+2H_2O(g)$       .......(3)

• Calculating the experimental yield of nitrogen gas:

By Stoichiometry of the reaction 3:

6 moles of NO is produced from 2 moles of $N_2O_4$

So, 0.383 moles of NO will be produced from = $\frac{2}{6}\times 0.383=0.128mol$ of $N_2O_4$

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 0.128 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 0.128=0.384mol$ 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:

$0.384mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.384mol\times 28g/mol)=10.75g$

• Calculating the theoretical yield of nitrogen gas:

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 1.11 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 1.11=3.33mol$ 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:

$3.33mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(3.33mol\times 28g/mol)=93.24g$

• To calculate the percentage yield of nitrogen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of nitrogen gas = 10.75 g

Theoretical yield of nitrogen gas = 93.24 g

Putting values in above equation, we get:

$\%\text{ yield of nitrogen gas}=\frac{10.75g}{93.24g}\times 100\\\\\% \text{yield of nitrogen gas}=11.53\%$

Hence, the percent yield of the nitrogen gas is 11.53 %.