Question

The reaction : $C{l}_2\left(g\right)+S_2{O}_3^{2-}⟶S{O}_4^{2-}+C{l}^{-}$ is to be carried out in basic medium. Starting with $0.15$ moles of $C{l}_2$, $0.010$ moles of $S_2{O}_3^{2-}$ and $0.30$ moles of $O{H}^{-}$, how many moles of $O{H}^{-}$ will be left in solution after the reaction is complete? Assume no other reaction occurs.

• A. $0.2$ mol
• B. $0.1$ mol
• C. $0.4$ mol
• D. $0.3$ mol

Answer 1

Answer: (A)

$0.2$ mol

If concentration of medium is desired after redox reaction, first balance the equation by ion-electron method and then solve it by mole concept.

Balanced equation is as follows:

$\begin{array}{ccccccccccc}4C{l}_2& +& S_2{O}_3^{2-}& +& 10O{H}^{-}& ⟶& 2S{O}_4^{2-}& +& 8C{l}^{-}& +& 5H_{2}O\\ & & & & & & & & & & \\ Molesbeforereaction& & & & & & & & & & \\ 0.15& & 0.01& & 0.30& & 0& & 0& & 0\\ & & & & & & & & & & \\ Molesafterreaction& & & & & & & & & & \\ & & & & & & & & & & \\ (0.15-4\times .01)& 0& (0.3-0.01\times 10)& & & & & & & & \\ & & & & & & & & & & \\ =0.11& 0& & & =0.2& & 0.02& & 0.08& & 0.05\end{array}$

$(\because$ Mole ratio for combinations is $C{l}_2:S_2{O}_3^{2-}:O{H}^{-}::4:1:10)$

$\therefore O{H}^{-}$ left after reaction is $0.2$ moles.