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Question: 0.2 mole of $Cl_2$ is reacted with 500 ml of 1.2 M - NaOH solution $Cl_2 + OH \rightarrow ClO_3^{-}...

0.2 mole of Cl2Cl_2 is reacted with 500 ml of 1.2 M - NaOH solution

Cl2+OH→ClO3−+Cl−+H2OCl_2 + OH \rightarrow ClO_3^{-} + Cl^{-} + H_2O

The incorrect information(s) related is/are -

A

Cl2Cl_2 is limiting reagent

B

0.4 mole NaOH will left unreacted

C

[ClO3−]finalsolution[ClO_3^{-}]_{final solution} = 0.133 M

D

[Cl−]finalsolution[Cl^{-}]_{final solution} = 0.267 M

Answer

B, D

Explanation

Solution

The given reaction is the disproportionation of chlorine in a basic medium, specifically to form chlorate and chloride ions. The unbalanced reaction provided is Cl2+OH→ClO3−+Cl−+H2OCl_2 + OH \rightarrow ClO_3^{-} + Cl^{-} + H_2O. This reaction needs to be balanced first.

1. Balancing the Chemical Equation:

This is a redox reaction where chlorine disproportionates (oxidation state changes from 0 to +5 in ClO3−ClO_3^{-} and to -1 in Cl−Cl^{-}). Let's balance it using the half-reaction method in basic medium.

  • Oxidation Half-Reaction: Cl2→ClO3−Cl_2 \rightarrow ClO_3^{-}

    • Balance Cl atoms: Cl2→2ClO3−Cl_2 \rightarrow 2ClO_3^{-}
    • Balance O atoms by adding H2OH_2O: Cl2+6H2O→2ClO3−Cl_2 + 6H_2O \rightarrow 2ClO_3^{-}
    • Balance H atoms by adding H+H^+: Cl2+6H2O→2ClO3−+12H+Cl_2 + 6H_2O \rightarrow 2ClO_3^{-} + 12H^+
    • Balance charge by adding electrons: Cl2+6H2O→2ClO3−+12H++10e−Cl_2 + 6H_2O \rightarrow 2ClO_3^{-} + 12H^+ + 10e^-

  • Reduction Half-Reaction: Cl2→Cl−Cl_2 \rightarrow Cl^{-}

    • Balance Cl atoms: Cl2→2Cl−Cl_2 \rightarrow 2Cl^{-}
    • Balance charge by adding electrons: Cl2+2e−→2Cl−Cl_2 + 2e^- \rightarrow 2Cl^{-}

  • Combine Half-Reactions: To balance electrons, multiply the reduction half-reaction by 5 (to get 10 electrons).

    • Oxidation: Cl2+6H2O→2ClO3−+12H++10e−Cl_2 + 6H_2O \rightarrow 2ClO_3^{-} + 12H^+ + 10e^-
    • Reduction: 5Cl2+10e−→10Cl−5Cl_2 + 10e^- \rightarrow 10Cl^{-}
    • Add them: 6Cl2+6H2O→2ClO3−+10Cl−+12H+6Cl_2 + 6H_2O \rightarrow 2ClO_3^{-} + 10Cl^{-} + 12H^+

  • Convert to Basic Medium: Add 12OH−12OH^- to both sides to neutralize 12H+12H^+.

    • 6Cl2+6H2O+12OH−→2ClO3−+10Cl−+12H2O6Cl_2 + 6H_2O + 12OH^- \rightarrow 2ClO_3^{-} + 10Cl^{-} + 12H_2O
    • Simplify H2OH_2O (subtract 6H2O6H_2O from both sides): 6Cl2+12OH−→2ClO3−+10Cl−+6H2O6Cl_2 + 12OH^- \rightarrow 2ClO_3^{-} + 10Cl^{-} + 6H_2O

  • Simplify Coefficients: Divide all coefficients by 2 to get the simplest whole number ratio.

    • 3Cl2+6OH−→ClO3−+5Cl−+3H2O3Cl_2 + 6OH^- \rightarrow ClO_3^{-} + 5Cl^{-} + 3H_2O

This is the correctly balanced chemical equation.

2. Calculate Initial Moles of Reactants:

  • Moles of Cl2=0.2Cl_2 = 0.2 mole
  • Volume of NaOH solution = 500 ml = 0.5 L
  • Molarity of NaOH solution = 1.2 M
  • Moles of NaOH = Molarity ×\times Volume = 1.2 mol/L×0.5 L=0.61.2 \text{ mol/L} \times 0.5 \text{ L} = 0.6 mole

3. Determine the Limiting Reagent:

From the balanced equation, 3 moles of Cl2Cl_2 react with 6 moles of OH−OH^- (NaOH). This is a 1:2 molar ratio (Cl2:NaOHCl_2 : NaOH).

  • For Cl2Cl_2 (0.2 mol):
    • Moles of NaOH required = 0.2 mol Cl2×6 mol NaOH3 mol Cl2=0.2×2=0.40.2 \text{ mol } Cl_2 \times \frac{6 \text{ mol } NaOH}{3 \text{ mol } Cl_2} = 0.2 \times 2 = 0.4 mol NaOH.
    • Available NaOH = 0.6 mol. Since 0.4 mol (required) < 0.6 mol (available), NaOH is in excess.
  • For NaOH (0.6 mol):
    • Moles of Cl2Cl_2 required = 0.6 mol NaOH×3 mol Cl26 mol NaOH=0.6×0.5=0.30.6 \text{ mol } NaOH \times \frac{3 \text{ mol } Cl_2}{6 \text{ mol } NaOH} = 0.6 \times 0.5 = 0.3 mol Cl2Cl_2.
    • Available Cl2Cl_2 = 0.2 mol. Since 0.2 mol (available) < 0.3 mol (required), Cl2Cl_2 is the limiting reagent.

Evaluation of Statements:

(A) Cl2Cl_2 is limiting reagent

Based on our calculation, Cl2Cl_2 is indeed the limiting reagent. This statement is CORRECT.

(B) 0.4 mole NaOH will left unreacted

  • Moles of NaOH reacted = Moles of Cl2Cl_2 reacted (limiting reagent) ×6 mol NaOH3 mol Cl2=0.2 mol Cl2×2=0.4\times \frac{6 \text{ mol } NaOH}{3 \text{ mol } Cl_2} = 0.2 \text{ mol } Cl_2 \times 2 = 0.4 mol NaOH.
  • Moles of NaOH initially present = 0.6 mol.
  • Moles of NaOH left unreacted = Initial moles - Reacted moles = 0.6 mol−0.4 mol=0.20.6 \text{ mol} - 0.4 \text{ mol} = 0.2 mol.

The statement says 0.4 mole NaOH will be left unreacted. This statement is INCORRECT.

(C) [ClO3−]finalsolution[ClO_3^{-}]_{final solution} = 0.133 M

  • Moles of ClO3−ClO_3^{-} produced = Moles of Cl2Cl_2 reacted ×1 mol ClO3−3 mol Cl2=0.2 mol Cl2×13=0.23\times \frac{1 \text{ mol } ClO_3^{-}}{3 \text{ mol } Cl_2} = 0.2 \text{ mol } Cl_2 \times \frac{1}{3} = \frac{0.2}{3} mol.
  • Final volume of solution = 0.5 L (assuming no significant volume change upon dissolution of Cl2Cl_2 gas).
  • Concentration of ClO3−=Moles of ClO3−Volume of solution=0.2/3 mol0.5 L=0.21.5=215≈0.1333ClO_3^{-} = \frac{\text{Moles of } ClO_3^{-}}{\text{Volume of solution}} = \frac{0.2/3 \text{ mol}}{0.5 \text{ L}} = \frac{0.2}{1.5} = \frac{2}{15} \approx 0.1333 M.

The statement says [ClO3−]finalsolution[ClO_3^{-}]_{final solution} = 0.133 M. This statement is CORRECT.

(D) [Cl−]finalsolution[Cl^{-}]_{final solution} = 0.267 M

  • Moles of Cl−Cl^{-} produced = Moles of Cl2Cl_2 reacted ×5 mol Cl−3 mol Cl2=0.2 mol Cl2×53=1.03\times \frac{5 \text{ mol } Cl^{-}}{3 \text{ mol } Cl_2} = 0.2 \text{ mol } Cl_2 \times \frac{5}{3} = \frac{1.0}{3} mol.
  • Final volume of solution = 0.5 L.
  • Concentration of Cl−=Moles of Cl−Volume of solution=1.0/3 mol0.5 L=1.01.5=1015=23≈0.6667Cl^{-} = \frac{\text{Moles of } Cl^{-}}{\text{Volume of solution}} = \frac{1.0/3 \text{ mol}}{0.5 \text{ L}} = \frac{1.0}{1.5} = \frac{10}{15} = \frac{2}{3} \approx 0.6667 M.

The statement says [Cl−]finalsolution[Cl^{-}]_{final solution} = 0.267 M. This statement is INCORRECT.

The incorrect information(s) are (B) and (D).