Chemistry - Solid State Question with Solution | TestHub

Let the initial moles of M be $n_M=1$ mol and O₂ be $n_{O_2}=0.8$ mol.

The stoichiometric reaction for M₂O formation is $2M+\frac{1}{2}{O}_2\to M_{2}O$.

If M is limiting, $1$ mol M reacts with $0.25$ mol O₂, leaving $0.8-0.25=0.55$ mol O₂ unreacted.

If O₂ is limiting, $0.8$ mol O₂ reacts with $3.2$ mol M, which is not possible as we only have $1$ mol M.

So, M is the limiting reactant, and $0.55$ mol O₂ remains unreacted.

The theoretical moles of M₂O formed is $0.5$ mol.

In the non-stoichiometric compound $M_{x}O$, there is a metal deficiency.

Let $y$ be the moles of M vacancies.

The actual moles of M in $M_{x}O$ is $1-y$.

The actual moles of O in $M_{x}O$ is $1$ mol (from $0.5$ mol O₂).

The ratio of moles of vacancies to remaining unreacted reactant is given as $5/7$.

So, $\frac{y}{0.55}=\frac{5}{7}$.

$y=\frac{5}{7}\times 0.55=\frac{2.75}{7}\approx 0.3928$ mol.

Number of vacancies $=y\times N_A=0.3928\times 6\times 10^{23}\approx 2.3568\times 10^{23}$.

Given number of vacancies $=N\times 10^{23}$.

Therefore, $N\approx 2.3568$. Rounding to the nearest integer, $N=2$.