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 ${\text{g mol}}^{-1}$) of $H,C$ and $O$ as $1,12$ and $16$, respectively).

The value of $y$ is

Answer 1

Molar mass of ${}^{\prime }{U}^{\prime }=122\text{g}$
$\underset{\left(0.1\text{mole}\right)}{P}\stackrel{100\text{\%}}{\to }\underset{\left(0.1\text{mole}\right)}{S}$

$\underset{\left(0.1\text{mole}\right)}{2S}\underset{\left(80\text{\%}\right)}{\overset{\text{Aldol condensation}}{\to }}\underset{(\frac{0.1}{2}\times 0.8)}{T}\stackrel{80\text{\%}}{\to }\underset{(\frac{0.1}{2}\times 0.8\times 0.8)}{U}$

Mass of ${}^{\prime }{U}^{\prime }=\frac{0.1}{2}\times 0.8\times 0.8\times 122=3.90\text{gm}$