What is the ratio of reducing agent to oxidizing agent , if the following reaction is correctly balanced ?
`NH_3 + O_2 rarr NO + H_2O`

Options

A. `4:5`
B. `5:4`
C. `5:3`
D. `3:5`

Explanation

To solve the problem of finding the ratio of the reducing agent to the oxidizing agent in the reaction \( \text{NH}_3 + \text{O}_2 \rightarrow \text{NO} + \text{H}_2\text{O} \), we will follow these steps:

### Step 1: Balance the Reaction
First, we need to balance the chemical equation.

1. **Count the number of atoms on each side**:
- Reactants:
- Nitrogen (N): 1 (from NH₃)
- Hydrogen (H): 3 (from NH₃)
- Oxygen (O): 2 (from O₂)
- Products:
- Nitrogen (N): 1 (from NO)
- Hydrogen (H): 2 (from H₂O)
- Oxygen (O): 1 (from NO) + 1 (from H₂O) = 2

2. **Balancing Hydrogen**:
We have 3 H in the reactants and 2 H in the products. To balance the hydrogen, we can multiply H₂O by 3/2:
\[ \text{NH}_3 + \text{O}_2 \rightarrow \text{NO} + \frac{3}{2} \text{H}_2\text{O} \]

3. **Balancing Nitrogen**:
Since we have 1 N in NH₃ and 1 N in NO, nitrogen is already balanced.

4. **Balancing Oxygen**:
Now we have 2 O from O₂ and 2 O from the products (1 from NO and 1.5 from H₂O). To balance the oxygen, we can multiply O₂ by 5/2:
\[ 2 \text{NH}_3 + \frac{5}{2} \text{O}_2 \rightarrow 2 \text{NO} + 3 \text{H}_2\text{O} \]

### Step 2: Identify Reducing and Oxidizing Agents
- **Reducing Agent**: A substance that donates electrons and is oxidized in the process. In this reaction, NH₃ is oxidized to NO, meaning it is the reducing agent.
- **Oxidizing Agent**: A substance that accepts electrons and is reduced in the process. In this reaction, O₂ is reduced to H₂O, meaning it is the oxidizing agent.

### Step 3: Determine the Ratio
From the balanced equation:
\[ 2 \text{NH}_3 + \frac{5}{2} \text{O}_2 \rightarrow 2 \text{NO} + 3 \text{H}_2\text{O} \]
- The stoichiometric coefficients show that for every 2 moles of NH₃ (reducing agent), there are \( \frac{5}{2} \) moles of O₂ (oxidizing agent).

To find the ratio of the reducing agent to the oxidizing agent:
\[ \text{Ratio} = \frac{\text{Moles of } \text{NH}_3}{\text{Moles of } \text{O}_2} = \frac{2}{\frac{5}{2}} = \frac{2 \times 2}{5} = \frac{4}{5} \]

### Final Answer
The ratio of the reducing agent (NH₃) to the oxidizing agent (O₂) is \( \frac{4}{5} \).

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