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 chemical reaction involved in the combustion of methane. The balanced chemical equation for the combustion of methane ($CH_4$) is:

$$ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O $$

From the equation, we see that one mole of methane produces two moles of water. Now, let's calculate the amount of water produced:

1. **Molar Mass Calculation:**
   - The molar mass of methane ($CH_4$) is approximately 16 g/mol.
   - The molar mass of water ($H_2O$) is approximately 18 g/mol.

2. **Moles of Methane:**
   - Given 16 g of methane, we have 1 mole of methane (since 16 g/mol is the molar mass).

3. **Moles of Water Produced:**
   - According to the balanced equation, 1 mole of methane produces 2 moles of water.

4. **Mass of Water Produced:**
   - 2 moles of water $\times$ 18 g/mol = 36 g of water.

Therefore, the combustion of 16 g of methane produces 36 g of water, which corresponds to Option 4.

**Explanation for Incorrect Options:**

- **Option 1 (8 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 less.

- **Option 2 (16 g):** This option is incorrect because it implies that only one mole of water is produced from one mole of methane. However, the balanced equation clearly shows that two moles of water are produced.

- **Option 3 (18 g):** This option is incorrect because it suggests that only one mole of water is produced. As per the balanced equation, two moles of water are produced from one mole of methane, resulting in 36 g of water, not 18 g.

In conclusion, the correct answer is Option 4 (36 g), as it accurately reflects the stoichiometry of the combustion reaction of methane.