Question

The reaction
$C{l}_2+S_2{O_3}^{2-}+{OH}^{-}⟶{{SO}_4}^{2-}+{Cl}^{-}+H_{2}O$
Starting with $0.15$ mole $C{l}_2,0.010$ mole$S_2{O_3}^{2-}$and $0.30$ mole ${OH}^{-}$, mole of $C{l}_2$ left in solution will be

• A. 0.11
• B. 0.01
• C. 0.04
• D. 0.09

Answer 1

Answer: (A)

0.11

${4Cl}_2+S_2{O_3}^{2-}+{10OH}^{-}⟶{{2SO}_4}^{2-}+{8Cl}^{-}+{5H}_{2}O$
limiting reagent is $S_2{O_3}^{2-}$
1 mole of $S_2{O_3}^{2-}$ react with 4 mole ${Cl}_2$
so,
$0.01$ mole $S_2{O_3}^{2-}=0.01\times 4=0.04mole{Cl}_2$
moles of ${Cl}_2$ left$=0.15-0.04=0.11$