`NO_(2)` required for a reaction is produced by the decomposition of `N_(2)O_(5)` is as per the equatiion `2N_(2)O_(5)(g)to4NO_(2)(g)+O_(2)(g)`. The initial concentration fo `N_(2)O_(5)` is `3.00 "mol"L^(-1)` and it is `2.75 "mol"L^(-1)` after 30 minutes. The rate of formation of `NO_(2)` is `1.67xx10^(-x) "mol" L^(-1)"min"^(-1)`. the numerical value of x is __________.

Explanation

To solve the problem, we need to determine the rate of formation of \( NO_2 \) from the decomposition of \( N_2O_5 \) based on the given reaction and concentration changes.

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
\[ 2 N_2O_5(g) \rightarrow 4 NO_2(g) + O_2(g) \]

2. **Identify initial and final concentrations**:
- Initial concentration of \( N_2O_5 \): \( [N_2O_5]_0 = 3.00 \, \text{mol/L} \)
- Final concentration of \( N_2O_5 \) after 30 minutes: \( [N_2O_5]_t = 2.75 \, \text{mol/L} \)

3. **Calculate the change in concentration of \( N_2O_5 \)**:
\[ \Delta [N_2O_5] = [N_2O_5]_0 - [N_2O_5]_t = 3.00 - 2.75 = 0.25 \, \text{mol/L} \]

4. **Determine the rate of decomposition of \( N_2O_5 \)**:
The rate of decomposition is given by:
\[ \text{Rate} = -\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t} \]
Here, \( \Delta t = 30 \, \text{minutes} = 30 \, \text{min} \).

Substituting the values:
\[ \text{Rate} = -\frac{1}{2} \frac{0.25 \, \text{mol/L}}{30 \, \text{min}} = -\frac{0.25}{60} = -0.004167 \, \text{mol/L/min} \]

5. **Convert to positive rate**:
Since we are interested in the rate of formation of \( NO_2 \), we take the positive value:
\[ \text{Rate of formation of } NO_2 = \frac{4}{2} \times \text{Rate of decomposition of } N_2O_5 = 2 \times 0.004167 = 0.008334 \, \text{mol/L/min} \]

6. **Express the rate in scientific notation**:
\[ 0.008334 \, \text{mol/L/min} = 8.334 \times 10^{-3} \, \text{mol/L/min} \]

7. **Compare with the given rate of formation**:
The problem states that the rate of formation of \( NO_2 \) is given as:
\[ 1.67 \times 10^{-x} \, \text{mol/L/min} \]
Setting the two expressions equal:
\[ 8.334 \times 10^{-3} = 1.67 \times 10^{-x} \]

8. **Solve for \( x \)**:
Rearranging gives:
\[ 8.334 = 1.67 \times 10^{3-x} \]
Dividing both sides by 1.67:
\[ \frac{8.334}{1.67} = 10^{3-x} \]
Calculating the left side:
\[ 4.99 \approx 5 \implies 10^{3-x} = 5 \]

Taking logarithm base 10:
\[ 3 - x = \log_{10}(5) \approx 0.699 \]
Thus,
\[ x \approx 3 - 0.699 \approx 2.301 \implies x \approx 2 \]

### Final Answer:
The numerical value of \( x \) is \( 2 \).