In redox reaction: Fe+HNO3→Fe(NO3)2+NH4NO3+H2O, what would be the coefficient of HNO3,Fe(NO3)2,NH4NO3
?

(A) 1:10:4

(B) 10:4:1

(C) 4:10:4

(D) 4:1:4

To solve this problem, we need to balance the redox reaction:

**Given reaction:**
\[ \text{Fe} + \text{HNO}_3 \rightarrow \text{Fe(NO}_3)_2 + \text{NH}_4\text{NO}_3 + \text{H}_2\text{O} \]

### Step 1: Identify the oxidation and reduction reactions
1. **Oxidation:**
\[ \text{Fe} \rightarrow \text{Fe}^{2+} \]
Here, Fe is oxidized from 0 to +2 oxidation state, losing 2 electrons.

2. **Reduction:**
Nitrogen in \(\text{HNO}_3\) (where the oxidation state of N is +5) is reduced to \(\text{NH}_4^+\) (where the oxidation state of N is -3). This is a gain of 8 electrons.

### Step 2: Balance the atoms and electrons for each half-reaction

**Oxidation half-reaction (Fe):**
\[
\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-
\]

**Reduction half-reaction (N):**
\[
\text{NO}_3^- + 8H^+ + 8e^- \rightarrow \text{NH}_4^+ + 2H_2O
\]

### Step 3: Equalize the number of electrons transferred
- The oxidation half-reaction transfers 2 electrons.
- The reduction half-reaction transfers 8 electrons.

To balance the electron transfer, multiply the oxidation half-reaction by 4:
\[
4\text{Fe} \rightarrow 4\text{Fe}^{2+} + 8e^-
\]

### Step 4: Combine the half-reactions
Now, combine the oxidation and reduction half-reactions:
\[
4\text{Fe} + 8\text{NO}_3^- + 8H^+ \rightarrow 4\text{Fe}^{2+} + \text{NH}_4^+ + 2H_2O
\]

### Step 5: Balance the full reaction
The balanced reaction is:
\[
4\text{Fe} + 10\text{HNO}_3 \rightarrow 4\text{Fe(NO}_3)_2 + \text{NH}_4\text{NO}_3 + 3H_2O
\]

### Step 6: Coefficients
From the balanced reaction, the coefficients of \( \text{HNO}_3 \), \( \text{Fe(NO}_3)_2 \), and \( \text{NH}_4\text{NO}_3 \) are:
\[
\text{HNO}_3 : \text{Fe(NO}_3)_2 : \text{NH}_4\text{NO}_3 = 10:4:1
\]

Thus, the correct answer is:

**(B) 10:4:1**