Ammonia is oxised by oxygen to give nitric oxide and water. The weight of water produced per gram of nitric oxide is `0.1 xx xg`. What is value of x.

Answer

To solve the problem, we need to determine the value of \( x \) in the expression for the weight of water produced per gram of nitric oxide. Let's break down the steps:

### Step 1: Write the balanced chemical equation
The reaction between ammonia (NH₃) and oxygen (O₂) to produce nitric oxide (NO) and water (H₂O) can be represented as:
\[ 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O} \]

### Step 2: Calculate the molar masses
- **Molar mass of nitric oxide (NO)**:
- Nitrogen (N) = 14 g/mol
- Oxygen (O) = 16 g/mol
- Molar mass of NO = 14 + 16 = 30 g/mol

- **Molar mass of water (H₂O)**:
- Hydrogen (H) = 1 g/mol (2 H = 2 g/mol)
- Oxygen (O) = 16 g/mol
- Molar mass of H₂O = 2 + 16 = 18 g/mol

### Step 3: Determine the mass ratio from the balanced equation
From the balanced equation, we see that:
- 4 moles of NO produce 6 moles of H₂O.

### Step 4: Calculate the mass of water produced per mass of nitric oxide
Using the molar masses:
- Mass of 4 moles of NO = \( 4 \times 30 \, \text{g} = 120 \, \text{g} \)
- Mass of 6 moles of H₂O = \( 6 \times 18 \, \text{g} = 108 \, \text{g} \)

Now, we can find the mass of water produced per gram of nitric oxide:
- Water produced per 120 g of NO = 108 g
- Therefore, water produced per 1 g of NO = \( \frac{108 \, \text{g}}{120 \, \text{g}} = 0.9 \, \text{g} \)

### Step 5: Set up the equation based on the problem statement
According to the problem, the weight of water produced per gram of nitric oxide is given as \( 0.1 \times x \) g. We can set up the equation:
\[ 0.1 \times x = 0.9 \]

### Step 6: Solve for \( x \)
To find \( x \):
\[ x = \frac{0.9}{0.1} = 9 \]

### Final Answer
The value of \( x \) is \( 9 \).

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