The amont of water produced by the combustion of 16 g of methane is

Options

A. 16 g
B. 36 g
C. 18 g
D. 32 g

Answer

To find the amount of water produced by the combustion of 16 g of methane (CH₄), we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of methane.
The balanced equation for the combustion of methane is:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]

### Step 2: Calculate the number of moles of methane.
To calculate the number of moles of methane, we use the formula:
\[ \text{Number of moles} = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}} \]

The molar mass of methane (CH₄) is calculated as follows:
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol × 4 = 4 g/mol
- Total molar mass of CH₄ = 12 + 4 = 16 g/mol

Now, substituting the values:
\[ \text{Number of moles of CH}_4 = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mole} \]

### Step 3: Use stoichiometry to find the moles of water produced.
From the balanced equation, we see that 1 mole of methane produces 2 moles of water. Therefore, if we have 1 mole of methane:
\[ \text{Moles of water produced} = 1 \text{ mole CH}_4 \times \frac{2 \text{ moles H}_2\text{O}}{1 \text{ mole CH}_4} = 2 \text{ moles H}_2\text{O} \]

### Step 4: Calculate the mass of water produced.
To find the mass of water produced, we again use the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]

The molar mass of water (H₂O) is:
- Hydrogen (H) = 1 g/mol × 2 = 2 g/mol
- Oxygen (O) = 16 g/mol
- Total molar mass of H₂O = 2 + 16 = 18 g/mol

Now, substituting the values:
\[ \text{Mass of water} = 2 \text{ moles} \times 18 \text{ g/mol} = 36 \text{ g} \]

### Final Answer:
The amount of water produced by the combustion of 16 g of methane is **36 g**.

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