What mass of carbon disulphide, $C{S}_2$ can be completely oxidized to $S{O}_2$ and $C{O}_2$ by the oxygen liberated when $325g$ of $N{a}_2{O}_2$ react with water?

The correct option is B $52.78g$
Given,

$325g$ of $N{a}_2{O}_2$ is reacted with water

Molar mass of $N{a}_2{O}_2=78g/mol$

$\therefore$ Number of moles of $N{a}_2{O}_2$ given=$\frac{325g}{78g/mol}$

$=4.166$ moles

The reaction involved is

$2N{a}_2{O}_2+2H_{2}O⟶4NaOH+O_2$

$2$ moles of $N{a}_2{O}_2$ gives $1$ moles of $O_2$

$\therefore 4.166$ moles of $N{a}_2{O}_2$ gives $⟶\frac{4.166}{2}$ moles of $O_2$

$=2.083$ moles of $O_2$

Now,

The molar mass of $C{S}_2=76/gmol$

The reaction involved id

$C{S}_2+3O_2⟶C{O}_2+2S{O}_2$

$1$ mole of $C{S}_2$ requires $3$ moles of $O_2$

Now, $2.083$ moles of $O_2$ will react with $x$ moles of $C{S}_2$ to produce $S{O}_2$ & $C{O}_2$

$x=0.6943$ moles of $C{S}_2$

$\therefore$ The mass of $C{S}_2$ which can be completely oxidized by $2.083$ moles of $O_2=0.6943mol\times 76g/mol$

$=52.76g$