Question

0.75 mol of solid ${{X}_{4}}$ and 2 mol of gaseous ${{O}_{2}}$ are heated to react completely in a sealed vessel to produce only one gaseous compound Y. After the compound is formed, the vessel is brought to the initial temperature, the pressure is found to be half of the initial pressure. The molecular formula of compound is:
(A)- ${{X}_{3}}{{O}_{4}}$
(B)- ${{X}_{2}}{{O}_{4}}$
(C)- ${{X}_{4}}{{O}_{4}}$
(D)- none of the above

Answer 1

Hint: To find the molecular formula of a compound, first determine the number of moles of the compound formed and then the number of atoms of each element in the compound.

Complete step by step solution:
The reaction of formation of compound Y can be written as follows:
\[{{X}_{4}}(s)+{{O}_{2}}(g)\xrightarrow{\Delta }Y(g)\]
Number of moles of ${{X}_{4}}$ reacted = 0.75
And, number of moles of ${{O}_{2}}$ reacted = 2
Let the pressure before heating the reaction mixture in the vessel be ${{P}_{1}}$. After the formation of the compound Y, the vessel is cooled down to its initial temperature. The final pressure ${{P}_{2}}$ was found half of the initial pressure ${{P}_{1}}$.i.e. ${{P}_{2}}=\dfrac{{{P}_{1}}}{2}$
Before heating the vessel, the pressure ${{P}_{1}}$ was only due to ${{O}_{2}}$ as ${{X}_{4}}$ is a solid. After all the reactant has reacted to form the gaseous compound Y, the final pressure ${{P}_{2}}$ is only due to compound Y.
Since, temperature remains the same before and after the reaction then the initial and final pressure are proportional to the number of moles of gaseous reactant and product, respectively such that
${{P}_{1}}\propto $ moles of ${{O}_{2}}$ and ${{P}_{2}}\propto $ moles of Y.
\[\begin{align}
  & \dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{moles\,of\,{{O}_{2}}}{moles\,of\,Y};\,\text{since}\,{{P}_{2}}=\dfrac{{{P}_{1}}}{2} \\
 & \Rightarrow moles\,of\,Y=\dfrac{2\times 1}{2}=1 \\
\end{align}\]
Thus, the number of moles of Y formed is 1. Now, one mole of Y has 0.75 mol of ${{X}_{4}}$ and 2 mol of ${{O}_{2}}$. We can calculate the number of atoms of X and O in Y as:
     \[0.75\,{{X}_{4}}:2\,{{O}_{2}}=0.75\times 4:2\times 2=3:4\]
The molecular formula of the compound Y is ${{X}_{3}}{{O}_{4}}$.

So, the correct answer is “Option A”.

Note: Pressure of the given system is only due to the reactant and product in gaseous phase. Reactant in the solid phase has no contribution to the pressure.