Question

During studies of the reaction below,

2 N2H4(l) + N2O4(l) ? 3 N2(g) + 4 H2O(g)
a chemical engineer measured a less-than-expected yield of N2 and discovered that the following side reaction occurs.
N2H4(l) + 2 N2O4(l) ? 6 NO(g) + 2 H2O(g)
In one experiment, 11.5 g of NO formed when 102.1 g of each reactant was used.
What is the highest percent yield of N2 that can be expected?

Answer 1

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

Step-by-step 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}=(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=(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 =
`(2)/(6)* 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 =
`(3)/(1)* 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* 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 =
`(3)/(1)* 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* 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}}* 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}=(10.75g)/(93.24g)* 100\\\\\% \text{yield of nitrogen gas}=11.53\%`

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