Ammonia and oxygen combine to produce water vapour and nitric oxide as per the chemical equation: `4NH_(3)(g) + 5O_(2)(g) to 6H_(2)O(g) + 4NO`.
(ii) How many moles of oxygen are required to burn 85 g of ammonia?

Explanation

To solve the problem of how many moles of oxygen are required to burn 85 g of ammonia (NH₃) according to the reaction:

\[ 4NH_3(g) + 5O_2(g) \rightarrow 6H_2O(g) + 4NO(g) \]

we will follow these steps:

### Step 1: Calculate the molar mass of ammonia (NH₃)

The molar mass of ammonia can be calculated using the atomic masses of nitrogen and hydrogen:
- Atomic mass of nitrogen (N) = 14 g/mol
- Atomic mass of hydrogen (H) = 1 g/mol

Since ammonia has one nitrogen atom and three hydrogen atoms, its molar mass is calculated as follows:

\[ \text{Molar mass of NH}_3 = 14 + (3 \times 1) = 14 + 3 = 17 \text{ g/mol} \]

### Step 2: Determine the number of moles of ammonia in 85 g

To find the number of moles of ammonia in 85 g, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Substituting the values:

\[ \text{Number of moles of NH}_3 = \frac{85 \text{ g}}{17 \text{ g/mol}} = 5 \text{ moles} \]

### Step 3: Use the stoichiometry of the reaction

From the balanced chemical equation, we see that 4 moles of ammonia react with 5 moles of oxygen. Therefore, we can set up a ratio to find out how many moles of oxygen are needed for 5 moles of ammonia:

\[ \frac{5 \text{ moles of } O_2}{4 \text{ moles of } NH_3} = x \text{ moles of } O_2 \text{ for } 5 \text{ moles of } NH_3 \]

Using cross-multiplication:

\[ x = \frac{5 \text{ moles of } O_2}{4 \text{ moles of } NH_3} \times 5 \text{ moles of } NH_3 \]

Calculating \(x\):

\[ x = \frac{5 \times 5}{4} = \frac{25}{4} = 6.25 \text{ moles of } O_2 \]

### Conclusion

Thus, the number of moles of oxygen required to burn 85 g of ammonia is **6.25 moles**.

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