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If ${Cl}_{2}$ and ${PCl}_{3}$ can react to form ${PCl}_{5}$ according to following reaction
${Cl}_{2} + {PCl}_{3} ⟶ {PCl}_{5}$
In tube A, 355 gm ${Cl}_{2}$ is taken and in tube B, 1375 gm ${PCl}_{3}$ is taken. When half the mass of gas in tube A is transferred to B, reaction took place in B. After the reaction is complete, half of the mass of mixture in B is again transferred to A to cause same reaction in container A. (no reversibility Select the correct option(s).

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Tube A contains 3.75 moles of ${PCl}_{5}$ and 1.25 moles of ${PCl}_{3}$ finally.

Tube A contains 2.5 moles of ${Cl}_{2}$ and 1.25 moles of ${PCl}_{5}$ finally

None of the above

Tube B contains 3.75 moles ${PCl}_{3}$ and 1.25 moles of ${PCl}_{5}$ finally .

answer is A, C.

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Detailed Solution

$\overset{(A)}{\underset{5 moles}{{Cl}_{2}}} + \underset{10moles}{\overset{(B)}{{PCl}_{3}}} \underset{}{} ⟶ {PCl}_{5}$

$\begin{array}{l} (B) {Cl}_{2} + {PCl}_{3} ⟶ {PCl}_{5} \\ \frac{5}{2} 10 \\ - 10 - \frac{5}{2} \frac{5}{2} \\ = \frac{15}{2} \end{array}$

${Cl}_{2} + {PCl}_{3} ⟶ {PCl}_{5}$

$\begin{array}{ccc} \frac{5}{2} & \frac{15}{4} & \frac{5}{2} \\ - & 1.25 & \frac{5}{4} + \frac{5}{2} = 3.75 \end{array}$

Moles of ${PCI}_{3}$ left in $B = \frac{15}{4} = 3.75$

Moles of ${PCl}_{5}$ left in $B = \frac{5}{4} = 1.25$

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