Dissolving 1.24 g of white phosphorus in boi | vedclass

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Question

(B) The reaction of white phosphorus $(P_4)$ with $NaOH$ is: $P_4 + 3NaOH + 3H_2O \longrightarrow PH_3 + 3NaH_2PO_2$ Moles of $P_4 = \frac{1.24 \ g}{1

Vedclass · Accepted Answer

$2.39$

Vedclass · Answer

(B) The reaction of white phosphorus $(P_4)$ with $NaOH$ is: $P_4 + 3NaOH + 3H_2O \longrightarrow PH_3 + 3NaH_2PO_2$ Moles of $P_4 = \frac{1.24 \ g}{124 \ g/mol} = 0.01 \ mol$. From the stoichiometry,$1 \ mol$ of $P_4$ produces $1 \ mol$ of $PH_3$ (gas $Q$). So,moles of $PH_3 = 0.01 \ mol$. The reaction of $PH_3$ with $CuSO_4$ is: $2PH_3 + 3CuSO_4 \longrightarrow Cu_3P_2 + 3H_2SO_4$ Moles of $CuSO_4$ required = $\frac{3}{2} 	imes 	ext{moles of } PH_3 = \frac{3}{2} 	imes 0.01 = 0.015 \ mol$. Molar mass of $CuSO_4 = 63 + 32 + (4 	imes 16) = 159 \ g/mol$. Mass of $CuSO_4 = 0.015 \ mol 	imes 159 \ g/mol = 2.385 \ g$. Rounding to two decimal places,the value is $2.39 \ g$.

Dissolving $1.24 \ g$ of white phosphorus 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 $H = 1, O = 16, Na = 23, P = 31, S = 32, Cu = 63$]

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