Question

Consider the reaction sequence from P to Q shown below. The overall yield of the major product Q from P is $75\%$. What is the amount in grams of Q obtained from $9.3mL$ of P? (Use density of $P=1.00gm{L}^{-1}$; Molar mass of $C=12.0$, $H=1.0$, $O=16.0$ and $N=14.0gmo{l}^{-1}$)

Answer 1

1. It is given that reactant P is aniline and the yield of product Q is $75\%$, amount of reactant P used is $9.3mL=9.3g$( as d=1.00g/mL).
2. The product formed in the given reaction sequence is as follows:
3. The molecular weight of aniline ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2=6\times 12+7\times 1+14=93$
4. It is also given that density of $P=1.00gm{L}^{-1}$.
5. Now, the number of moles of P, $n_p=\frac{9.3}{93}=0.1moles$
6. The mole ratio of ${\mathrm{PhNH}}_2:{{\mathrm{PhN}}_2}^{+}:{\mathrm{PhPhN}}_2\mathrm{PhOH}=1:1:1$
7. So, when the extent of reaction is $100\%$ the number of moles of Q formed is $0.1moles$, but for $75\%$ yield the number of moles of Q formed is $n_Q=0.1\times \frac{75}{100}=0.075moles$.
8. Molecular weight of product Q, ${\mathrm{PhPhN}}_2\mathrm{PhOH}=6\times 12+12\times 1+16+14\times 2=248\mathrm{g}$.
9. So the amount of product $$\begin{gathered} Q=0.075\times 248g \\ =18.6g\approx 19g \end{gathered}$$
10. Therefore, amount of product $\mathit{Q}\mathbf{=}\mathbf{19}\mathbf{}\mathit{g}$.