When hydrogen burns in oxygen, water vapour is produced:
`2H_(2)(g) + O_(2)(g) to 2H_(2)O` (vapour).
How many moles of steam is obtained from `0.5` mol of `O_(2)` used?

Explanation

To solve the problem, we need to analyze the chemical reaction and determine how many moles of steam (water vapor) are produced from a given amount of oxygen.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The balanced equation for the combustion of hydrogen in oxygen is:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \]
This equation tells us that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water vapor.

2. **Identify the Molar Ratios**:
From the balanced equation, we can see the molar ratio between oxygen and water vapor:
- 1 mole of \(O_2\) produces 2 moles of \(H_2O\).

3. **Calculate the Moles of Water Vapor from Given Moles of Oxygen**:
We are given 0.5 moles of \(O_2\). Using the molar ratio:
- For every 1 mole of \(O_2\), we get 2 moles of \(H_2O\).
- Therefore, for 0.5 moles of \(O_2\):
\[ \text{Moles of } H_2O = 0.5 \, \text{moles of } O_2 \times \frac{2 \, \text{moles of } H_2O}{1 \, \text{mole of } O_2} = 1 \, \text{mole of } H_2O \]

4. **Conclusion**:
Thus, when 0.5 moles of \(O_2\) are used, 1 mole of steam (water vapor) is produced.

### Final Answer:
**1 mole of steam is obtained from 0.5 moles of \(O_2\) used.**