The numbe of moles of oxygen obtained by the electrolytic decomposition of 108 g water is

Options

A. 2.5
B. 3
C. 5
D. 7.5

Explanation

To determine the number of moles of oxygen obtained by the electrolytic decomposition of 108 g of water, we can follow these steps:

### Step 1: Calculate the number of moles of water (H₂O) in 108 g.
- The molar mass of water (H₂O) is calculated as follows:
- Hydrogen (H) has a molar mass of approximately 1 g/mol, and there are 2 hydrogen atoms in water.
- Oxygen (O) has a molar mass of approximately 16 g/mol, and there is 1 oxygen atom in water.

Therefore, the molar mass of water is:
\[ \text{Molar mass of H₂O} = (2 \times 1) + 16 = 18 \text{ g/mol} \]

- Now, we can calculate the number of moles of water in 108 g:
\[ \text{Number of moles of H₂O} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{108 \text{ g}}{18 \text{ g/mol}} = 6 \text{ moles} \]

### Step 2: Use the stoichiometry of the decomposition reaction to find moles of oxygen (O₂).
- The balanced chemical equation for the electrolysis of water is:
\[ 2 \text{H₂O} \rightarrow 2 \text{H₂} + \text{O₂} \]
- From the equation, we see that 2 moles of water produce 1 mole of oxygen. Therefore, we can set up a ratio:
\[ \text{From 2 moles of H₂O} \rightarrow 1 mole of O₂ \]
\[ \text{From 6 moles of H₂O} \rightarrow \frac{6}{2} \text{ moles of O₂} = 3 \text{ moles of O₂} \]

### Final Answer:
The number of moles of oxygen obtained by the electrolytic decomposition of 108 g of water is **3 moles**.

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