Number of moles of oxygen liberated by electrolysis of 90g of water

Options

A. 9 moles
B. 4-5 moles
C. 2.5moles
D. 5 moles.

Explanation

To find the number of moles of oxygen liberated by the electrolysis of 90 grams of water, we can follow these steps:

### Step 1: Calculate the number of moles of water
The number of moles can be calculated using the formula:

\[ \text{Number of moles} = \frac{\text{Weight of substance (g)}}{\text{Molecular weight (g/mol)}} \]

For water (H₂O), the molecular weight is approximately 18 g/mol.

Given:
- Weight of water = 90 g
- Molecular weight of water = 18 g/mol

Substituting the values:

\[ \text{Number of moles of water} = \frac{90 \text{ g}}{18 \text{ g/mol}} = 5 \text{ moles} \]

### Step 2: Determine the moles of oxygen produced
From the electrolysis of water, the balanced chemical equation is:

\[ 2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + \text{O}_2 \]

This equation shows that 2 moles of water produce 1 mole of oxygen. Therefore, we can set up a ratio:

\[ \text{Moles of O}_2 = \frac{1 \text{ mole O}_2}{2 \text{ moles H}_2\text{O}} \times \text{Moles of H}_2\text{O} \]

Substituting the number of moles of water we calculated:

\[ \text{Moles of O}_2 = \frac{1}{2} \times 5 = 2.5 \text{ moles} \]

### Conclusion
The number of moles of oxygen liberated by the electrolysis of 90 grams of water is **2.5 moles**.

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