Dissolving 1 . 24   g of white phosphorous in boiling NaOH solution in an inert atmosphere gives a gas Q . The amount of CuSO 4 (in g ) required to completely consume the gas Q is____[Given: Atomic mass of H = 1 ,   O = 16 ,   Na = 23 ,   P = 31 ,   S = 32 ,   Cu = 63 ]
A. A
P 4 1 . 24   g + 3 NaOH + 3 H 2 O → PH 3 + 3 NaH 2 PO 2 Number of moles of P 4   =   Given   mass molar   mass   =   1 . 24 124   =   0 . 01   moles As NaOH is present in excess. So, amount of phosphine formed is 0 . 01 mole (as P 4 is limiting reagent) 2 PH 3 0 . 01 mole + 3 CuSO 4 → 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 = 3 × 0 . 01 2 mole Mass of CuSO 4 (in g ) required = 0 . 03 2 × 63 + 32 + 16 × 4 = 0 . 03 2 × 159 = 2 . 38   g
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