Dissolving 1 . 24g of white phosphorous in boiling NaOH solution in...

$\mathop{{P}_{4}}_{1 . 24g} + 3NaOH + 3{H}_{2}O \to {PH}_{3} + 3{NaH}_{2}{PO}_{2}$

Number of moles of ${P}_{4} = \dfrac{Givenmass}{molarmass} = \dfrac{1 . 24}{124} = 0 . 01moles$

As $NaOH$ is present in excess. So, amount of phosphine formed is $0 . 01$ mole (as ${P}_{4}$ is limiting reagent)

$\mathop{2{PH}_{3}}_{0 . 01mole} + 3{CuSO}_{4} \to {Cu}_{3}{P}_{2} + 3{H}_{2}{SO}_{4}$

Since 2 moles of phosphine required 3 moles of ${CuSO}_{4}$.

Amount of ${CuSO}_{4}$ required $= \dfrac{3 \times 0 . 01}{2}$ mole

Mass of ${CuSO}_{4}$ (in $g$) required $= \dfrac{0 . 03}{2} \times \left(63 + 32 + 16 \times 4\right)$

$= \dfrac{0 . 03}{2} \times 159$

$= 2 . 38g$