Chemistry - Complete P-block Question with Solution | TestHub

$\underset{1.24 \mathrm{g}}{{\mathrm{P}}_4}+3\mathrm{NaOH}+3{\mathrm{H}}_2\mathrm{O}\to {\mathrm{PH}}_3+3{\mathrm{NaH}}_2{\mathrm{PO}}_2$

Number of moles of ${\mathrm{P}}_4 = \frac{\mathrm{Given} \mathrm{mass}}{\mathrm{molar} \mathrm{mass}} = \frac{1.24}{124} = 0.01 \mathrm{moles}$

As $\mathrm{NaOH}$ is present in excess. So, amount of phosphine formed is $0.01$ mole (as ${\mathrm{P}}_4$ is limiting reagent)

$\underset{0.01\mathrm{mole}}{2{\mathrm{PH}}_3}+3{\mathrm{CuSO}}_4\to {\mathrm{Cu}}_3{\mathrm{P}}_2+3{\mathrm{H}}_2{\mathrm{SO}}_4$

Since 2 moles of phosphine required 3 moles of ${\mathrm{CuSO}}_4$.

Amount of ${\mathrm{CuSO}}_4$ required $=\frac{3\times 0.01}{2}$ mole

Mass of ${\mathrm{CuSO}}_4$ (in $\mathrm{g}$) required $=\frac{0.03}{2}\times \left(63+32+16\times 4\right)$

$=\frac{0.03}{2}\times 159$

$=2.38 \mathrm{g}$