Number of moles of KI required to prepare 0.4 moles of `K_(2)Hgl_(4)`, when `HgCl_(2)` reacts with `Kl`

Options

A. `"0.4 moles of KI"`
B. `"0.8 moles of KI"`
C. `"1.6 moles of KI"`
D. `"3.2 moles of KI"`

Explanation

To find the number of moles of KI required to prepare 0.4 moles of K₂HgI₄, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**: The reaction between mercury(II) chloride (HgCl₂) and potassium iodide (KI) can be represented as:
\[ \text{HgCl}_2 + 4 \text{KI} \rightarrow \text{K}_2\text{HgI}_4 + 2 \text{KCl} \]
From this equation, we see that 1 mole of K₂HgI₄ is produced from 4 moles of KI.

2. **Determine the Moles of KI Needed for 1 Mole of K₂HgI₄**: According to the balanced equation, to produce 1 mole of K₂HgI₄, we need 4 moles of KI.

3. **Calculate the Moles of KI Required for 0.4 Moles of K₂HgI₄**:
- If 1 mole of K₂HgI₄ requires 4 moles of KI, then 0.4 moles of K₂HgI₄ will require:
\[ \text{Moles of KI} = 4 \text{ moles of KI} \times 0.4 \text{ moles of K}_2\text{HgI}_4 \]
\[ \text{Moles of KI} = 1.6 \text{ moles of KI} \]

4. **Final Answer**: Therefore, the number of moles of KI required to prepare 0.4 moles of K₂HgI₄ is **1.6 moles**.

### Summary:
To prepare 0.4 moles of K₂HgI₄, you need **1.6 moles of KI**.