The moles of `O_(2)` required for reacting with `6*8` g ammonia
`(...NH_(3)+....O_(2)to....NO+....H_(2)O)` is

Options

A. `5`
B. `2*5`
C. `1`
D. `0*5`

Explanation

To solve the problem of determining the moles of \( O_2 \) required to react with \( 6.8 \, \text{g} \) of ammonia \( (NH_3) \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The first step is to write the balanced chemical equation for the reaction between ammonia and oxygen. The unbalanced equation is:
\[ NH_3 + O_2 \rightarrow NO + H_2O \]
To balance it, we find that:
\[ 4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O \]

### Step 2: Calculate the molar mass of ammonia
Next, we need to calculate the molar mass of ammonia \( (NH_3) \):
- Nitrogen (N) has a molar mass of approximately \( 14 \, \text{g/mol} \).
- Hydrogen (H) has a molar mass of approximately \( 1 \, \text{g/mol} \).
- Therefore, the molar mass of \( NH_3 \) is:
\[ 14 + (3 \times 1) = 17 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of ammonia
Now, we can calculate the number of moles of ammonia in \( 6.8 \, \text{g} \):
\[ \text{Moles of } NH_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{6.8 \, \text{g}}{17 \, \text{g/mol}} \approx 0.4 \, \text{mol} \]

### Step 4: Use the stoichiometry of the balanced equation
From the balanced equation, we see that \( 4 \) moles of \( NH_3 \) react with \( 5 \) moles of \( O_2 \). We can set up a ratio to find the moles of \( O_2 \) required for \( 0.4 \, \text{mol} \) of \( NH_3 \):
\[ \frac{5 \, \text{moles of } O_2}{4 \, \text{moles of } NH_3} = \frac{x \, \text{moles of } O_2}{0.4 \, \text{moles of } NH_3} \]
Cross-multiplying gives us:
\[ x = \frac{5}{4} \times 0.4 = 0.5 \, \text{mol} \]

### Step 5: Conclusion
Thus, the moles of \( O_2 \) required for reacting with \( 6.8 \, \text{g} \) of ammonia is \( 0.5 \, \text{mol} \).

### Final Answer
The moles of \( O_2 \) required is \( 0.5 \, \text{mol} \).

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