Question

For the following reaction scheme, percentage yields are given along the arrow:$\text{x g}$ and $\text{y g}$ are the mass of $R$ and $U$, respectively.
(Use : Molar mass (in ${gmol}^{-1}$) of $H,C$ and $O$ as $1,12$ and $16$, respectively)

The value of $x$ is:

Answer 1

Molar mass of $P=40$
So moles of $P=\frac{4}{40}=0.1\text{mol}$
$\underset{\left(0.1\text{mol}\right)}{P}\stackrel{75\text{\%}}{\to }\underset{(0.1\times \frac{3}{4})}{Q}$

$\underset{(0.1\times \frac{3}{4})}{3Q}\stackrel{40\text{\%}}{\to }\underset{(\frac{1}{3}\times 0.1\times \frac{3}{4}\times 0.4)}{R}$

So, moles of $R=0.01$ mole
Molar mass of $\left(R\right)=162$
So, $x=0.01\times 162=1.62\text{g}$