Nitric acid is manufactured by the Ostwald process, in which
nitrogen dioxide reacts with water.
`3NO_(2)(g) + H_(2)O(l) to 2HNO_(3) (aq) + NO(g)`
How many grams of nitrogen dioxide are required in this reaction to produce 25.2 gm `HNO_(3)`?

Answer

To solve the problem of how many grams of nitrogen dioxide (NO₂) are required to produce 25.2 grams of nitric acid (HNO₃) using the given reaction, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the Ostwald process is:
\[ 3 \text{NO}_2 (g) + \text{H}_2\text{O} (l) \rightarrow 2 \text{HNO}_3 (aq) + \text{NO} (g) \]

### Step 2: Calculate the molar mass of HNO₃
The molar mass of HNO₃ can be calculated as follows:
- Hydrogen (H): 1 g/mol
- Nitrogen (N): 14 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol

Thus, the molar mass of HNO₃ is:
\[ 1 + 14 + 48 = 63 \text{ g/mol} \]

### Step 3: Calculate the number of moles of HNO₃ produced
To find the number of moles of HNO₃ produced from 25.2 grams, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

So,
\[ \text{Number of moles of HNO}_3 = \frac{25.2 \text{ g}}{63 \text{ g/mol}} \approx 0.4 \text{ moles} \]

### Step 4: Use stoichiometry to find the moles of NO₂ required
From the balanced equation, we see that 2 moles of HNO₃ are produced from 3 moles of NO₂. Therefore, we can set up a ratio:
\[ \frac{3 \text{ moles NO}_2}{2 \text{ moles HNO}_3} \]

Using this ratio, we can find the moles of NO₂ required for 0.4 moles of HNO₃:
\[ \text{Moles of NO}_2 = 0.4 \text{ moles HNO}_3 \times \frac{3 \text{ moles NO}_2}{2 \text{ moles HNO}_3} = 0.6 \text{ moles NO}_2 \]

### Step 5: Calculate the mass of NO₂ required
Now, we need to calculate the mass of NO₂ required. The molar mass of NO₂ is:
- Nitrogen (N): 14 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol

Thus, the molar mass of NO₂ is:
\[ 14 + 32 = 46 \text{ g/mol} \]

Now, we can find the mass of NO₂:
\[ \text{Mass of NO}_2 = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass of NO}_2 = 0.6 \text{ moles} \times 46 \text{ g/mol} = 27.6 \text{ g} \]

### Final Answer
Therefore, the mass of nitrogen dioxide required to produce 25.2 grams of HNO₃ is **27.6 grams**.

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