The amount of water produced by the combustion of 16 g of methane is :

Question

Accepted Answer

To determine the amount of water produced by the combustion of 16 g of methane, we need to consider the balanced chemical equation for the combustion of methane:

$$ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} $$

From this equation, we can see that one mole of methane ($\text{CH}_4$) produces two moles of water ($\text{H}_2\text{O}$).

1. **Molar Mass Calculation:**
   - The molar mass of methane ($\text{CH}_4$) is approximately 16 g/mol.
   - The molar mass of water ($\text{H}_2\text{O}$) is approximately 18 g/mol.

2. **Combustion of 16 g of Methane:**
   - 16 g of methane is equivalent to 1 mole of methane.
   - According to the balanced equation, 1 mole of methane produces 2 moles of water.

3. **Mass of Water Produced:**
   - Since 1 mole of water weighs 18 g, 2 moles of water will weigh $2 \times 18 = 36$ g.

Therefore, the combustion of 16 g of methane produces 36 g of water, which corresponds to Option 3.

**Explanation for Incorrect Options:**

- **Option 1 (9 g):** This option is incorrect because it suggests that less than one mole of water is produced, which contradicts the stoichiometry of the balanced equation. One mole of methane produces two moles of water, not half a mole.

- **Option 2 (18 g):** This option is incorrect because it represents the mass of only one mole of water. The balanced equation shows that two moles of water are produced from one mole of methane.

- **Option 4 (72 g):** This option is incorrect because it suggests that four moles of water are produced, which is double the actual amount according to the balanced equation. The correct stoichiometry indicates that only two moles of water are produced from one mole of methane.

In conclusion, the correct answer is Option 3, as it accurately reflects the stoichiometry of the combustion reaction of methane, resulting in the production of 36 g of water.