Chemistry - Hydrocarbon Question with Solution | TestHub

${\mathrm{Mg}}_2{\mathrm{C}}_3\stackrel{{\mathrm{H}}_2\mathrm{O}}{\to }\mathrm{Mg}{\left(\mathrm{OH}\right)}_2+\underset{\mathrm{Propyne} \left(4\mathrm{g}\right)}{{\mathrm{C}}_3{\mathrm{H}}_4\left(\mathrm{P}\right)}$

${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH}\underset{\begin{array}{l}33\mathrm{K}\\ \left(100\mathrm{\%}\right)\end{array}}{\overset{{\mathrm{Hg}}^{2+}}{\to }}{\mathrm{CH}}_3-\underset{\underset{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-\underset{\left(\mathrm{S}\right)}{{\mathrm{CH}}_3}$

Now, one moe of propyne give one mole of acetone. $\left(100\%\right)$. one mole of acetone give $0.5$ moles of $\mathrm{\alpha },\mathrm{\beta }$-unsaturated

compound $\left[100\% \mathrm{yield}\right]$. But given that $80\%$ yield.

No. of moles of $\mathrm{\alpha },\mathrm{\beta }$-unsaturated compound $=0.04 \mathrm{mol}$. $=0.04 \mathrm{mol}$ of $\left(\mathrm{T}\right)$ can give $=0.04 \mathrm{mol}$ of $\left(\mathrm{U}\right)$ $\left[100\% \mathrm{yield}\right]$.

But given that $80\%$ yield.

No. of moles of $\left(\mathrm{U}\right)$ formed $=0.032$

mass of $\left(\mathrm{U}\right)$ formed $0.032\times 100$

$=3.20$