Question

$C{l}_2\left(g\right)+S_2{O}_3^{2-}⟶S{O}_4^{2-}+C{l}^{-}+S$

If $50.0$ mL of $0.01MN{a}_2{S}_2{O}_3$ solution and $5\times {10}^{-4}$ moles of $C{l}_2$ react according to the given equation then how many millimoles of the reducing agent are left?

• A. $0.5$
• B. $0.05$
• C. $0.1$
• D. $0$

Answer 1

Answer: (D)

$0$

Moles of $S_2{O}_3^{2-}$ $=0.1\times 0.05=0.0005$
Moles of $C{l}_2$ $=0.0005$

$C{l}_2+2e^{-}\to 2C{l}^{-}$

$S_2{O}_3^{2-}\to S{O}_4^{2-}+S+2e^{-}$

As for a combined reaction, $2$ electrons are transferred and the moles of the reactants are equal.
$\therefore$ all the reducing agent, $S_2{O}_3^{2-}$ is used up.

$\therefore$ no millimoles of reducing agent are left behind in the reaction.

Hence, the correct option is $\left(D\right)$.