When water is electrolysed, hydrogen and oxygen gas are produced. If 1.008 g of `H_(2)` is liberated at the cathode. What mass of `O_(2)` is formed at the anode?

Explanation

To solve the problem of determining the mass of oxygen gas (O₂) produced when 1.008 g of hydrogen gas (H₂) is liberated at the cathode during the electrolysis of water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction:**
The electrolysis of water can be represented by the following half-reactions:
- At the cathode (reduction):
\[ 2H^+ + 2e^- \rightarrow H_2 \quad \text{(1 mole of H₂ is produced from 2 moles of H⁺)} \]
- At the anode (oxidation):
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \quad \text{(1 mole of O₂ is produced from 2 moles of H₂O)} \]

2. **Calculate Moles of Hydrogen Gas (H₂):**
The molar mass of hydrogen gas (H₂) is approximately 2 g/mol.
To find the number of moles of H₂ produced:
\[ \text{Moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2} = \frac{1.008 \text{ g}}{2 \text{ g/mol}} = 0.504 \text{ moles} \]

3. **Determine Moles of Oxygen Gas (O₂) Produced:**
From the stoichiometry of the reactions, we know that for every 2 moles of H₂ produced, 1 mole of O₂ is produced. Thus:
\[ \text{Moles of } O_2 = \frac{0.504 \text{ moles of } H_2}{2} = 0.252 \text{ moles} \]

4. **Calculate the Mass of Oxygen Gas (O₂):**
The molar mass of oxygen gas (O₂) is approximately 32 g/mol. Now, we can calculate the mass of O₂ produced:
\[ \text{Mass of } O_2 = \text{moles of } O_2 \times \text{molar mass of } O_2 = 0.252 \text{ moles} \times 32 \text{ g/mol} = 8.064 \text{ g} \]

5. **Final Answer:**
Therefore, the mass of oxygen gas (O₂) formed at the anode is approximately:
\[ \text{Mass of } O_2 \approx 8 \text{ g} \]

### Summary:
The mass of oxygen gas produced when 1.008 g of hydrogen gas is liberated during the electrolysis of water is approximately **8 g**.