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 $P =1.00 \,g \,mL ^{-1}$; Molar mass of $C =12.0, H =1.0, O =16.0$ and $N =14.0 \,g \,mol ^{-1}$ )

Answer 1

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 gm / mol

Moles of P =Moles of Q=$\frac{9.3}{93}=0.1$
So, amount of (Q) =$0.1 \times 248 \times 0.75 = 18.60$ gram