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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 112.4 g of each reactant was used. What is the highest percent yield of N2 that can be expected?

Respuesta :

Answer: The percent yield of the nitrogen gas is 10.5 %.

Explanation:

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

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

  • For [tex]N_2O_4[/tex] :

Given mass of [tex]N_2O_4[/tex] = 112.4 g

Molar mass of [tex]N_2O_4[/tex] = 92 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }N_2O_4=\frac{102.1g}{92g/mol}=1.22mol[/tex]

  • For NO:

Given mass of NO = 11.5 g

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of NO}=\frac{11.5g}{30g/mol}=0.383mol[/tex]

For the given chemical reactions:

[tex]2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)[/tex]      ......(2)

[tex]N_2H_4(l)+2N_2O_4(l)\rightarrow 6NO(g)+2H_2O(g)[/tex]       .......(3)

  • Calculating the theoretical yield of nitrogen gas:

By Stoichiometry of the reaction 2:

1 mole of [tex]N_2O_4[/tex] produces 3 moles of nitrogen gas

So, 1.22 moles of [tex]N_2O_4[/tex] will produce = [tex]\frac{3}{1}\times 1.22=3.66mol[/tex] of nitrogen gas

Now, calculating the theoretical yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 3.66 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

[tex]3.66mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(3.66mol\times 28g/mol)=102.48g[/tex]

  • Calculating the experimental yield of nitrogen gas:

By Stoichiometry of the reaction 3:

6 moles of NO is produced from 2 moles of [tex]N_2O_4[/tex]

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

By Stoichiometry of the reaction 2:

1 mole of [tex]N_2O_4[/tex] produces 3 moles of nitrogen gas

So, 0.128 moles of [tex]N_2O_4[/tex] will produce = [tex]\frac{3}{1}\times 0.128=0.384mol[/tex] 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:

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

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

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield of nitrogen gas = 10.75 g

Theoretical yield of nitrogen gas = 102.48 g

Putting values in above equation, we get:

[tex]\%\text{ yield of nitrogen gas}=\frac{10.75g}{102.48g}\times 100\\\\\% \text{yield of nitrogen gas}=10.5\%[/tex]

Hence, the percent yield of the nitrogen gas is 10.5 %.

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