NO reacts with `O_(2)` to form `NO_(2)`. When 10 g of `NO_(2)` is formed during the reaction, the mass of `O_(2)` consumed is

Options

A. `1.90 g`
B. `5.0 g`
C. `3.48 g`
D. 13.9 g

Explanation

To solve the problem, we need to determine the mass of \( O_2 \) consumed when 10 g of \( NO_2 \) is formed from the reaction between \( NO \) and \( O_2 \).

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
The reaction between nitrogen monoxide (\( NO \)) and oxygen (\( O_2 \)) to form nitrogen dioxide (\( NO_2 \)) can be represented as:
\[ 2 NO + O_2 \rightarrow 2 NO_2 \]

2. **Determine the molar mass of \( NO_2 \)**:
The molar mass of \( NO_2 \) can be calculated as follows:
- Molar mass of Nitrogen (N) = 14 g/mol
- Molar mass of Oxygen (O) = 16 g/mol
Therefore, the molar mass of \( NO_2 \) is:
\[ 14 + (2 \times 16) = 14 + 32 = 46 \text{ g/mol} \]

3. **Calculate the number of moles of \( NO_2 \) produced**:
To find the number of moles of \( NO_2 \) formed from 10 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Thus,
\[ \text{Number of moles of } NO_2 = \frac{10 \text{ g}}{46 \text{ g/mol}} \approx 0.217 \text{ moles} \]

4. **Use stoichiometry to find the moles of \( O_2 \) consumed**:
From the balanced equation, we see that 1 mole of \( O_2 \) produces 2 moles of \( NO_2 \). Therefore, the moles of \( O_2 \) consumed can be calculated as:
\[ \text{Moles of } O_2 = \frac{0.217 \text{ moles of } NO_2}{2} \approx 0.1085 \text{ moles} \]

5. **Calculate the mass of \( O_2 \) consumed**:
Now, we can find the mass of \( O_2 \) using its molar mass (32 g/mol):
\[ \text{Mass of } O_2 = \text{Number of moles} \times \text{Molar mass} \]
Thus,
\[ \text{Mass of } O_2 = 0.1085 \text{ moles} \times 32 \text{ g/mol} \approx 3.48 \text{ g} \]

### Final Answer:
The mass of \( O_2 \) consumed is approximately **3.48 g**.