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 9.3 $mL$ of $P$?
(Use density of $P$ = 1.00 $gm{L}^{-1}$; Molar mass of $C=12.0$, $H=1.0$, $O=16.0$ and $N=14.0gmo{l}^{-1}$

Answer 1

Molecular weight of aniline (C₆H₅NH₂) = 77 + 14 + 2 = 93

Density of $P$ = 1 $g/mL$

d = m/v

or m = dv

Mass of $P$ = 9.3 × 1 = 9.3 $g$

9.3 mL of $P$ = 9.3 $g$

The mole ratio $PhN{H}_2:Ph{N}_2^{+}:$
= 1:1:1

So,the mole of $Q$ formed will be 0.1 mole and extent of reaction is 100 % but if it is 75 % yield.

Then amount of $Q$ = 0.1 x 75/100 = 0.075 mol

The molecular formula of $Q$ = C₁₆H₁₂ON₂

So, M. wt. of $Q$ = 16 x 12 + 12 x 1 + 16 + 2 x 14

= 192 + 12 + 16 + 28

= 248 $g/mol$

So, amount of $Q$ = 248 x 0.075 = 18.6 $g$