In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can obtained starting only with `10.00 g` of ammonia and `20.00 g` of oxygen?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation.
The balanced chemical reaction for the oxidation of ammonia (NH₃) by oxygen (O₂) to produce nitric oxide (NO) and water (H₂O) is:

\[ 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O} \]

### Step 2: Calculate the molar masses of the reactants and products.
- Molar mass of NH₃ = 14 (N) + 3 (H) = 17 g/mol
- Molar mass of O₂ = 2 × 16 = 32 g/mol
- Molar mass of NO = 14 (N) + 16 (O) = 30 g/mol

### Step 3: Convert the masses of the reactants to moles.
- Moles of NH₃ in 10.00 g:
\[ \text{Moles of NH}_3 = \frac{10.00 \text{ g}}{17 \text{ g/mol}} \approx 0.588 \text{ mol} \]

- Moles of O₂ in 20.00 g:
\[ \text{Moles of O}_2 = \frac{20.00 \text{ g}}{32 \text{ g/mol}} = 0.625 \text{ mol} \]

### Step 4: Determine the stoichiometric ratios from the balanced equation.
From the balanced equation, the stoichiometric ratio is:
- 4 moles of NH₃ react with 5 moles of O₂.

### Step 5: Find the limiting reagent.
Using the stoichiometric ratios, we can determine how much NH₃ is needed for the available O₂:
- For 0.625 moles of O₂, the amount of NH₃ required:
\[ \text{Required NH}_3 = 0.625 \text{ mol O}_2 \times \frac{4 \text{ mol NH}_3}{5 \text{ mol O}_2} = 0.5 \text{ mol NH}_3 \]

Since we have 0.588 moles of NH₃ available, and only 0.5 moles are needed, O₂ is the limiting reagent.

### Step 6: Calculate the amount of NO produced.
From the balanced equation, 5 moles of O₂ produce 4 moles of NO. Therefore, for 0.625 moles of O₂:
\[ \text{Moles of NO produced} = 0.625 \text{ mol O}_2 \times \frac{4 \text{ mol NO}}{5 \text{ mol O}_2} = 0.5 \text{ mol NO} \]

### Step 7: Convert moles of NO to grams.
Now, we convert the moles of NO produced to grams:
\[ \text{Mass of NO} = 0.5 \text{ mol NO} \times 30 \text{ g/mol} = 15 \text{ g} \]

### Final Answer:
The maximum weight of nitric oxide (NO) that can be obtained is **15 g**.

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