Question

During studies of the reaction in Sample Problem

$\mathbf{2}{\mathit{N}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{4}}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{+}{\mathit{N}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{\to }\mathbf{3}{\mathit{N}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

a chemical engineer measured a less-than-expected yield of N2 and discovered that the following side reaction occurs:

${\mathit{N}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{4}}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{N}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{\to }\mathbf{6}\mathit{N}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

In one experiment, 10.0 g of NO formed when 100.0 g of each reactant was used. What is the highest percent yield of N2 that can be expected?

Short answer

The highest percent yield of N₂ is 89.8%.

Step-by-step solution

Finding the moles

The molar ratios can be based only on the chemical equations.

$$\begin{gathered} Moles of N_2(from N_2{H}_4)=100 g N_2{H}_4\times \frac{1 mol N_2{H}_4}{32.05 g N_2{H}_4}\times \frac{3 mol N_2}{2 mol N_2{H}_4} \\ =4.65 mol N_2. \\ Moles of N_2(from N_2{O}_4)=100 g N_2{O}_4\times \frac{1 mol N_2{O}_4}{92.01 g N_2{O}_4}\times \frac{3 mol N_2}{1 mol N_2{O}_4} \\ =3.26 mol N_2. \\ Moles of N_2\left(theoretical\right)=3.26 mol N_2\times \frac{28.02 g N_2}{1 mol N_2} \\ =91.3 g N_2. \end{gathered}$$

Finding the highest percent yield

The actual yield can be calculated as:

$$\begin{gathered} Mass of N_2{O}_4(for NO)=10 g of NO\times \frac{1 mol NO}{30.01 g NO}\times \frac{2 mol N_2{O}_4}{6 mol NO}\times \frac{92.01 g N_2{O}_4}{1 mol N_2{O}_4} \\ =10.2 g N_2{O}_4. \\ Mass of N_2{O}_4(for N_2)=100 g-10.2 g \\ =89.8 N_2{O}_4. \\ Mass of N_2\left(actual\right)=89.8 g N_2{O}_4\times \frac{1 mol N_2{O}_4}{92.01 g N_2{O}_4}\times \frac{3 mol N_2}{1 mol N_2{O}_4}\times \frac{28.02 g N_2}{1 mol N_2} \end{gathered}$$

On calculating the percent yield, we get:

$$\begin{gathered} \% yield of N_2=\frac{82 g N_2}{91.3 g N_2}\times 100 \\ =89.8\%. \end{gathered}$$