`2KI+I_(2)+22HNO_(3)rarr2HIO_(3)+2KIO_(3)+22NO_(2)+10H_(2)O`
If 3 mole of `KI & 2` moles `I_(2)` are reacted with excess of `HNO_(3)`. Volume of `NO_(2)` gas evolved at `NTP` is `:`
`2KI+I_(2)+22HNO_(3)rarr2HIO_(3)+2KIO_(3)+22NO_(2)+10H_(2)O`

Options

A. `1075.2Lt`
B. `739.2 L t`
C. `44.8 L t`
D. `67.2 L t`

Explanation

To solve the problem, we need to analyze the given chemical reaction and determine how much NO₂ gas is produced when 3 moles of KI and 2 moles of I₂ are reacted with an excess of HNO₃.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
The balanced equation is:
\[ 2KI + I_2 + 22HNO_3 \rightarrow 2HIO_3 + 2KIO_3 + 22NO_2 + 10H_2O \]

2. **Identify the stoichiometry of the reaction:**
From the balanced equation, we can see that:
- 2 moles of KI produce 22 moles of NO₂.
- Therefore, 1 mole of KI produces \( \frac{22}{2} = 11 \) moles of NO₂.

3. **Calculate the moles of NO₂ produced from 3 moles of KI:**
If 1 mole of KI produces 11 moles of NO₂, then:
\[ 3 \text{ moles of KI} \rightarrow 3 \times 11 = 33 \text{ moles of NO₂} \]

4. **Calculate the volume of NO₂ at NTP:**
At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of 33 moles of NO₂ is:
\[ \text{Volume} = 33 \text{ moles} \times 22.4 \text{ L/mole} = 739.2 \text{ liters} \]

5. **Final answer:**
The volume of NO₂ gas evolved at NTP is **739.2 liters**.