$92 \mathrm{g}$ mixture of ${\mathrm{CaCO}}_3$, and ${\mathrm{MgCO}}_3$ heated strongly in an open vessel. After complete decomposition of the carbonates it was found that the weight of residue left behind is $48 \mathrm{g}$. Find the mass of ${\mathrm{MgCO}}_3$ in grams in the mixture.

 

If $0.5$ mole of silver salt is taken and weight of residue obtained is $216\mathrm{g}.(\mathrm{Ag}=108\mathrm{g}/\mathrm{mol})$. Then which the following is correct:

(i) ${\mathrm{K}}_4\mathrm{Fe}(\mathrm{CN}{)}_6+3{\mathrm{H}}_2{\mathrm{SO}}_4⟶2{\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{FeSO}}_4+6\mathrm{HCN}$

(ii) $6\mathrm{HCN}+12{\mathrm{H}}_2\mathrm{O}⟶6\mathrm{HCOOH}+6{\mathrm{NH}}_3$

(iii) (a) $6{\mathrm{NH}}_3+3{\mathrm{H}}_2{\mathrm{SO}}_4⟶3{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$
(b) $6\mathrm{HCOOH}\stackrel{{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }6\mathrm{CO}+6{\mathrm{H}}_2\mathrm{O}$

Above steps of reactions occur in a container starting with one ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right],5$ mole of ${\mathrm{H}}_2{\mathrm{SO}}_4$ and enough water. Find out the limiting reagent in step (i) and calculate maximum moles of $\mathrm{CO}$ gas and ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ that can be produced.