Question

$100 \text{ mL}$ of $0.05 \text{ M } \text{Cu}^{2+}$ aqueous solution is added to $1 \text{ L}$ of $0.1 \text{ M } \text{KI}$ solution. The number of moles of $\text{I}_2$ and $\text{Cu}_2\text{I}_2$ formed are respectively

• A. $2.5\times 10^{-3}, 5\times 10^{-3}$
• B. $5\times 10^{-3}, 5\times 10^{-3}$
• C. $5\times 10^{-3}, 2.5\times 10^{-3}$
• D. $2.5\times 10^{-3}, 2.5\times 10^{-3}$

Hint:

For reactions involving stoichiometry, the first step is always to write and balance the chemical equation. Then, use the mole ratio to identify the limiting reagent, which controls the amount of product formed. The final calculation is based on the limiting reagent.

Answer 1

The Correct Option is D

Step 1: Calculate the initial moles of the reactants.
The number of moles is $n = M \times V$ (in Liters). Moles of $\text{Cu}^{2+}$: $n_{\text{Cu}^{2+}} = (0.05 \text{ M}) \times (0.1 \text{ L}) = 0.005 \text{ moles} = 5 \times 10^{-3} \text{ moles}$. Moles of $\text{KI}$ (which provides $\text{I}^-$): $n_{\text{I}^-} = (0.1 \text{ M}) \times (1.0 \text{ L}) = 0.1 \text{ moles} = 100 \times 10^{-3} \text{ moles}$.

Step 2: Write the balanced redox reaction equation.
$\text{Cu}^{2+}$ is reduced by $\text{I}^-$ (iodide) to $\text{Cu}^{+}$ (which precipitates as $\text{Cu}_2\text{I}_2$), and $\text{I}^-$ is oxidized to $\text{I}_2$. \[ 2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^{-}(\text{aq}) \to \text{Cu}_2\text{I}_2(\text{s}) + \text{I}_2(\text{aq}). \]

Step 3: Identify the limiting reagent.
From the stoichiometry of the reaction, $2$ moles of $\text{Cu}^{2+}$ require $4$ moles of $\text{I}^-$. The required mole ratio is $\frac{\text{I}^-}{\text{Cu}^{2+}} = 2$. Available mole ratio: $\frac{n_{\text{I}^-}}{n_{\text{Cu}^{2+}}} = \frac{0.1 \text{ mol}}{0.005 \text{ mol}} = 20$. Since the available ratio (20) is much greater than the required ratio (2), $\text{Cu}^{2+}$ is the limiting reagent and will be completely consumed.

Step 4: Calculate the moles of products formed.
Based on the limiting reagent $\text{Cu}^{2+}$ ($5 \times 10^{-3} \text{ moles}$): From the balanced equation: Moles of $\text{Cu}_2\text{I}_2$: $\frac{n_{\text{Cu}_2\text{I}_2}}{\text{coefficient}} = \frac{n_{\text{Cu}^{2+}}}{\text{coefficient}}$. \[ n_{\text{Cu}_2\text{I}_2} = \frac{1}{2} n_{\text{Cu}^{2+}} = \frac{1}{2} (5 \times 10^{-3} \text{ moles}) = 2.5 \times 10^{-3} \text{ moles}. \] Moles of $\text{I}_2$: \[ n_{\text{I}_2} = \frac{1}{2} n_{\text{Cu}^{2+}} = 2.5 \times 10^{-3} \text{ moles}. \]

Step 5: Conclude the final answer.
The number of moles of $\text{I}_2$ and $\text{Cu}_2\text{I}_2$ formed are $2.5\times 10^{-3}$ moles and $2.5\times 10^{-3}$ moles respectively.