In acidic medium Zn reduces nitrate ion ot `NH_(4)^(+)` ion according to the reaction `underset("(unbalanced)")(Zn+NO_(3)^(-))rarrZn^(2+)+NH_(4)^(+)+H_(2)O`
How many moles of `HCl` are required to reduce half a mole of `NaNO_(3)` completely? Assume the availability of sufficient Zn.

Options

A. 5
B. 4
C. 3
D. 2

Answer

To solve the problem, we need to determine how many moles of HCl are required to reduce half a mole of NaNO3 completely in acidic medium using zinc.

### Step-by-Step Solution:

1. **Understand the Reaction**:
The unbalanced reaction given is:
\[ \text{Zn} + \text{NO}_3^- \rightarrow \text{Zn}^{2+} + \text{NH}_4^+ + \text{H}_2\text{O} \]
In acidic medium, the nitrate ion (NO3-) is reduced to ammonium ion (NH4+).

2. **Balance the Reaction**:
To balance the reaction, we need to account for the atoms and charges on both sides. The balanced equation is:
\[ 4 \text{Zn} + 10 \text{H}^+ + 4 \text{NO}_3^- \rightarrow 4 \text{Zn}^{2+} + 4 \text{NH}_4^+ + 6 \text{H}_2\text{O} \]
This tells us that 4 moles of Zn react with 10 moles of H+ (which comes from HCl) and 4 moles of NO3-.

3. **Determine H+ Requirement for NaNO3**:
Since NaNO3 dissociates into Na+ and NO3-, we can see that:
- 1 mole of NaNO3 provides 1 mole of NO3-.
- From the balanced reaction, 4 moles of NO3- require 10 moles of H+.

4. **Calculate H+ from HCl**:
Therefore, for 1 mole of NaNO3:
\[ 1 \text{ mole of NO}_3^- \text{ requires } \frac{10}{4} = 2.5 \text{ moles of H}^+ \]
Since H+ comes from HCl, we need 2.5 moles of HCl for 1 mole of NaNO3.

5. **Find Requirement for Half a Mole of NaNO3**:
For half a mole of NaNO3:
\[ \frac{1}{2} \text{ mole of NaNO}_3 \text{ requires } 2.5 \times \frac{1}{2} = 1.25 \text{ moles of HCl} \]

6. **Final Calculation**:
Thus, the number of moles of HCl required to reduce half a mole of NaNO3 completely is:
\[ 1.25 \text{ moles of HCl} \]

### Conclusion:
The answer is that **1.25 moles of HCl** are required to reduce half a mole of NaNO3 completely.