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 using zinc in an acidic medium.

### Step-by-Step Solution:

1. **Understand the Reaction**:
The unbalanced reaction given is:
\[ \text{Zn} + \text{NO}_3^{-} \rightarrow \text{Zn}^{2+} + \text{NH}_4^{+} + \text{H}_2\text{O} \]
This indicates that zinc reduces nitrate ions to ammonium ions in an acidic medium.

2. **Balancing the Reaction**:
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation. The balanced reaction is:
\[ 8\text{H}^+ + \text{Zn} + 2\text{NO}_3^{-} \rightarrow \text{Zn}^{2+} + 2\text{NH}_4^{+} + 4\text{H}_2\text{O} \]
From this balanced equation, we see that 1 mole of Zn reacts with 2 moles of NO₃⁻ to produce 2 moles of NH₄⁺.

3. **Determine Moles of NaNO₃**:
We are given that we need to reduce half a mole of NaNO₃. Since NaNO₃ dissociates into Na⁺ and NO₃⁻, half a mole of NaNO₃ corresponds to half a mole of NO₃⁻:
\[ \text{Moles of NO}_3^{-} = 0.5 \text{ moles} \]

4. **Using the Balanced Equation**:
From the balanced equation, we can see that 2 moles of NO₃⁻ require 8 moles of H⁺ (from HCl) for complete reduction:
\[ 2 \text{ moles NO}_3^{-} \rightarrow 8 \text{ moles H}^+ \]
Therefore, for 0.5 moles of NO₃⁻:
\[ \text{Moles of H}^+ = \frac{8}{2} \times 0.5 = 2 \text{ moles} \]

5. **Calculating Moles of HCl**:
Since HCl provides H⁺ ions in a 1:1 ratio, the moles of HCl required will be equal to the moles of H⁺ needed:
\[ \text{Moles of HCl} = 2 \text{ moles} \]

### Final Answer:
Thus, **2 moles of HCl** are required to reduce half a mole of NaNO₃ completely.