When 3 moles of Cl ₂( g ) are passed into 2 L of hot and concentrated 5 M NaOH solution, a disproportionation reaction occurs. Assuming the volume of the solution remains constant, the final concentrations of Cl ⁻ , ClO ₃⁻ and unreacted OH

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When 3 moles of Cl ₂( g ) are passed into 2 L of hot and concentrated 5 M NaOH solution, a disproportionation reaction occurs. Assuming the volume of the solution remains constant, the final concentrations of Cl ⁻ , ClO ₃⁻ and unreacted OH ⁻ respectively are

Quantrex Academy · Accepted Answer

Correct answer: C. 2.5 M , 0.5 M , 2.0 M

In hot and concentrated alkali, chlorine undergoes disproportionation as follows: 3 Cl ₂ + 6 OH ⁻ 5 Cl ⁻ + ClO ₃⁻ + 3 H ₂ O Initial moles of Cl ₂ = 3 mol Initial moles of OH ⁻ = 2 L 5 M = 10 mol From the stoichiometry, 3 moles of Cl ₂ react with 6 moles of OH ⁻ . Thus, Cl ₂ is the limiting reagent. Moles of Cl ⁻ formed = 5 mol Moles of ClO ₃⁻ formed = 1 mol Moles of unreacted OH ⁻ = 10 - 6 = 4 mol Since the volume is 2 L , the final concentrations are: [ Cl ⁻] = 5 2 = 2.5 M [ ClO ₃⁻] = 1 2 = 0.5 M [ OH ⁻] = 4 2 = 2

When 3 moles of Cl ₂( g ) are passed into 2 L of hot and concentrated 5 M NaOH solution, a disproportionation reaction occurs. Assuming the volume of the solution remains constant, the final concentrations of Cl ⁻ , ClO ₃⁻ and unreacted OH ⁻ respectively are

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