Chemistry - Complete d-block Question with Solution | TestHub

The amount of energy released during preparation of a green crystalline complex using $FeC{l}_3,KOH$ and ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ in the mole ratio $1:6:3$ respectively is $\mathbf{X}\left(\frac{\mathrm{kcal}}{\mathrm{mol}}\right)$.

If same reaction is carried out using same amount of ${\mathrm{FeCl}}_3$ as before but now mass ratio of ${\mathrm{FeCl}}_3$, KOH & ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ is $1:6:3$ respectively, then the energy released is $\mathbf{Y}\left(\frac{\mathrm{kcal}}{\mathrm{mol}}\right)$.

Find the value of($\mathrm{A}+\mathrm{B}$);

(where, $\mathbf{A}=\left(\frac{\mathrm{X}}{\mathrm{Y}}\right)$ and $\mathbf{B}=$ Number of unpaired electron(s) in green crystalline complex.) $[\mathrm{Fe}=56,\mathrm{Cl}=35.5,\mathrm{C}=12,\mathrm{O}=16,\mathrm{K}=39,\mathrm{H}=1]$