In the commercial manufacture of nitric acid, how many moles of `NO_(2)` produce 7.33 mol of `HNO_(3)` in the reaction :
`3NO_(2)(g)+H_(2)O(l) rarr 2HNO_(3)(aq)+NO(g)`?

Explanation

To find out how many moles of \( NO_2 \) are needed to produce 7.33 moles of \( HNO_3 \) in the reaction:

\[ 3NO_2(g) + H_2O(l) \rightarrow 2HNO_3(aq) + NO(g) \]

we can use stoichiometry based on the coefficients from the balanced chemical equation.

### Step 1: Identify the stoichiometric ratio

From the balanced equation, we can see that:

- 3 moles of \( NO_2 \) produce 2 moles of \( HNO_3 \).

### Step 2: Set up the ratio

We can set up a ratio to find out how many moles of \( NO_2 \) are required for 7.33 moles of \( HNO_3 \):

\[ \frac{3 \text{ moles } NO_2}{2 \text{ moles } HNO_3} \]

### Step 3: Calculate moles of \( NO_2 \)

Let \( x \) be the number of moles of \( NO_2 \) needed to produce 7.33 moles of \( HNO_3 \):

\[ \frac{x \text{ moles } NO_2}{7.33 \text{ moles } HNO_3} = \frac{3 \text{ moles } NO_2}{2 \text{ moles } HNO_3} \]

Now, we can cross-multiply to solve for \( x \):

\[ x \cdot 2 = 3 \cdot 7.33 \]

\[ 2x = 21.99 \]

\[ x = \frac{21.99}{2} = 10.995 \]

### Step 4: Conclusion

Thus, approximately 11.00 moles of \( NO_2 \) are required to produce 7.33 moles of \( HNO_3 \).

### Final Answer

**11.00 moles of \( NO_2 \)** are needed to produce **7.33 moles of \( HNO_3 \)**.

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