JEE Advance - Chemistry (2022 - Paper 1 Online - No. 5)
Dissolving $1.24 \mathrm{~g}$ of white phosphorous in boiling $\mathrm{NaOH}$ solution in an inert atmosphere gives a gas $\mathbf{Q}$. The amount of $\mathrm{CuSO}_{4}$ (in g) required to completely consume the gas $\mathbf{Q}$ is _________.
[Given: Atomic mass of $\mathrm{H}=1, \mathrm{O}=16, \mathrm{Na}=23, \mathrm{P}=31, \mathrm{~S}=32, \mathrm{Cu}=63$ ]
Explanation
$$\mathop {{P_4}}\limits_{\matrix{ {1.24\,g} \cr {or} \cr {0.01\,mole} \cr } } +3NaOH+3H_2O\to PH_3+3NaH_2PO_2$$
As NaOH is present in excess. So, amount of phosphine formed is 0.01 mole (as P4 is limiting)
$$\mathop {2P{H_3}}\limits_{0.01\,mole} +3CuSO_4\to Cu_3P_2+3H_2SO_4$$
Amount of CuSO4 required = $$\frac{3\times0.01}{2}$$ mole
Mass of CuSO4 (in g) required = $$\frac{0.03}{2}\times(63+32+16\times4)$$
$$=\frac{0.03}{2}\times159=2.38$$ g
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