Mass of Copper Sulfate Required to Consume Phosphine Gas

To find the amount of $\text{CuSO}_4$ required to completely consume the gas $\mathbf{Q}$, we follow a step-by-step stoichiometric analysis:

Step 1: Reaction of white phosphorus with boiling $\text{NaOH}$ When white phosphorus ($\text{P}_4$) is heated with an aqueous solution of $\text{NaOH}$ in an inert atmosphere, it undergoes disproportionation to produce phosphine gas ($\text{PH}_3$), which is gas $\mathbf{Q}$, and sodium hypophosphite ($\text{NaH}_2\text{PO}_2$):

$\text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow \text{PH}_3(\text{g}) + 3\text{NaH}_2\text{PO}_2$

Given:

• Mass of white phosphorus ($\text{P}_4$) = $1.24\text{ g}$
• Molar mass of $\text{P}_4 = 4 \times 31 = 124\text{ g/mol}$

$\text{Moles of }\text{P}_4 = \frac{1.24\text{ g}}{124\text{ g/mol}} = 0.01\text{ mol}$

From the balanced equation, $1\text{ mole of }\text{P}_4$ produces $1\text{ mole of }\text{PH}_3$: $\text{Moles of gas }\mathbf{Q}\text{ }(\text{PH}_3) = 0.01\text{ mol}$

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Step 2: Reaction of phosphine gas ($\text{PH}_3$) with $\text{CuSO}_4$ Phosphine gas reacts with aqueous copper sulfate solution to form copper phosphide precipitate ($\text{Cu}_3\text{P}_2$) according to the balanced equation:

$3\text{CuSO}_4 + 2\text{PH}_3 \rightarrow \text{Cu}_3\text{P}_2 + 3\text{H}_2\text{SO}_4$

From the stoichiometry of this reaction: $2\text{ moles of }\text{PH}_3 \text{ require } 3\text{ moles of }\text{CuSO}_4$

Thus, the number of moles of $\text{CuSO}_4$ required is: $\text{Moles of }\text{CuSO}_4 = \frac{3}{2} \times \text{Moles of }\text{PH}_3 = \frac{3}{2} \times 0.01 = 0.015\text{ mol}$

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Step 3: Calculation of the mass of $\text{CuSO}_4$ Using the given atomic masses ($\text{Cu} = 63, \text{S} = 32, \text{O} = 16$):

$\text{Molar mass of }\text{CuSO}_4 = 63 + 32 + (4 \times 16) = 159\text{ g/mol}$

Now, calculating the required mass of $\text{CuSO}_4$: $\text{Mass of }\text{CuSO}_4 = 0.015\text{ mol} \times 159\text{ g/mol} = 2.385\text{ g}$

Answer: The amount of $\text{CuSO}_4$ required is $2.385\text{ g}$ (or $2.38$ to $2.39$).