How many oxygen atoms are present in $2$ moles of Plaster of Paris $\left(\mathrm{POP}\right)$?

How many liters of ${\mathrm{CO}}_2$ at $\mathrm{STP }$will be formed when $0.01\mathrm{mol}$  of  ${\mathrm{H}}_2{\mathrm{SO}}_4$reacts with excess of ${\mathrm{Na}}_2{\mathrm{CO}}_3$?

${\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

A solution containing $2.675 \mathrm{g}$ of ${\mathrm{CoCl}}_3.6{\mathrm{NH}}_3 \left(\mathrm{molar} \mathrm{mass} =267.5\mathrm{g} {\mathrm{mol}}^{-1}\right)$ is passed through a cation exchanger. The chloride ions obtained in the solution were treated with an excess of ${\mathrm{AgNO}}_3$ to give $4.31 \mathrm{g}$ of $\mathrm{AgCl}$ $(\mathrm{molar} \mathrm{mass}=143.5 \mathrm{g} {\mathrm{mol}}^{-1})$. The formula of the complex is: $\left(\mathrm{Ag}=108\mathrm{u}\right)$

$\text{ Hint:molesof }{\mathrm{Cl}}^{-}=\mathrm{moles}\text{ of }\mathrm{AgCl}=\frac{4.31}{143.5}=0.03.$

$\because 0.01\mathrm{mol} \mathrm{of} {\mathrm{CoCl}}_3.6{\mathrm{NH}}_3 \mathrm{produces} 0.03 \mathrm{mol} \mathrm{of} {\mathrm{Cl}}^{-}$

$\therefore 1\text{ mole of }{\mathrm{CoCl}}_3·6{\mathrm{NH}}_3\text{ contains }3{\mathrm{Cl}}^{-}\text{anions }$