KOH + Cl₂ → Cl⁻ + ClO⁻ + K⁺

KOH = 2 M, 2 L
Cl₂ = 1 mol

Find the concentration of each product and choose the correct option.

To find the correct concentration of each product, we first start with the chemical reaction equation representing the reaction of potassium hydroxide (KOH) with chlorine gas (Cl₂):
2KOH + Cl₂ → KCl + KClO + H₂O
In ionic form, since KOH, KCl, and KClO dissociate completely in water:
2K⁺ + 2OH⁻ + Cl₂ → K⁺ + Cl⁻ + K⁺ + ClO⁻ + H₂O
Which simplifies to the net ionic equation:
2OH⁻ + Cl₂ → Cl⁻ + ClO⁻ + H₂

Let us calculate the initial moles of the reactants:
1. Moles of KOH:
$\text{Moles of KOH}=\text{Molarity}\times \text{Volume (L)}=2\text{M}\times 2\text{L}=4\text{mol}$
This also gives us:
Initial moles of OH⁻ = 4 mol
Initial moles of K⁺ = 4 mol

2. Moles of Cl₂:
Given as 1 mol.

Now, we identify the limiting reactant:
According to the balanced equation, 2 moles of OH⁻ react with 1 mole of Cl₂.
To react 1 mole of Cl₂ completely, we need:
$2\times 1\text{mol}=2\text{mol of}{\text{OH}}^{-}$
Since we have 4 moles of OH⁻ available, OH⁻ is in excess, and Cl₂ is the limiting reactant.

Next, we calculate the moles of the products formed from the 1 mole of limiting reactant Cl₂:
According to the stoichiometry of the reaction:
1 mole of Cl₂ produces 1 mole of Cl⁻ and 1 mole of ClO⁻.
Therefore, at the end of the reaction:
Moles of Cl⁻ formed = 1 mol
Moles of ClO⁻ formed = 1 mol

Since the volume of the solution is 2 L, we can find the concentration (molarity) of each ion product:
$\left[{\text{Cl}}^{-}\right]=\frac{1\text{mol}}{2\text{L}}=0.5\text{M}$
$\left[{\text{ClO}}^{-}\right]=\frac{1\text{mol}}{2\text{L}}=0.5\text{M}$

Thus, the concentrations of Cl⁻ and ClO⁻ are both 0.5 M.
Therefore, the correct option is:
[Cl⁻] = [ClO⁻] = 0.5 M