$K_{\mathrm{c}}$ for the following reaction is 99.0
$$ A_2(g) \stackrel{T(K)}{\rightleftharpoons} B_2(g) $$
In a one litre flask, 2 moles of $A_2$ was heated to $T(\mathrm{~K})$ and the above equilibrium is reached. The concentration at equilibrium of $A_2$ and $B_2$ are $C_1\left(A_2\right)$ and $C_2\left(B_2\right)$ respectively. Now, one mole of $A_2$ was added to flask and heated to $T(\mathrm{~K})$ to established the equilibrium again. The concentration of $A_2$ and $B_2$ are $C_3\left(A_2\right)$ and $C_4\left(B_2\right)$ respectively. what is the value of $C_3\left(A_2\right)$ in $\mathrm{mol} \mathrm{L}^{-1}$ ?
The correct statements among the following are
(i) saline hydrides produce $\mathrm{H}_2$ gas when reacted with water
(ii) presently $\sim 77 \%$ of the industrial dihydrogen is produced from coal
(iii) commercially marketed $\mathrm{H}_2 \mathrm{O}_2$ contains $3 \% \mathrm{H}_2 \mathrm{O}_2$