Sure, let's calculate the amount of water produced by the combustion of 16 grams of methane (CH₄).
First, we need to write down the balanced chemical equation for the combustion of methane:
[tex]\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \][/tex]
This equation tells us that 1 mole of methane (CH₄) reacts with 2 moles of oxygen (O₂) to produce 1 mole of carbon dioxide (CO₂) and 2 moles of water (H₂O).
Next, we need to determine the molar masses of the involved substances:
- Molar mass of methane (CH₄) = 16 g/mol (12 g/mol for carbon + 4 g/mol for hydrogen)
- Molar mass of water (H₂O) = 18 g/mol (16 g/mol for oxygen + 2 g/mol for hydrogen)
Now, we'll calculate the moles of methane:
[tex]\[ \text{Moles of CH}_4 = \frac{\text{Mass of CH}_4}{\text{Molar mass of CH}_4} = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mol} \][/tex]
According to the balanced chemical equation, 1 mole of methane will produce 2 moles of water. Therefore:
[tex]\[ \text{Moles of H}_2\text{O} = 2 \times \text{Moles of CH}_4 = 2 \times 1 \text{ mol} = 2 \text{ mol} \][/tex]
Finally, we calculate the mass of water produced from the moles of water:
[tex]\[ \text{Mass of H}_2\text{O} = \text{Moles of H}_2\text{O} \times \text{Molar mass of H}_2\text{O} = 2 \text{ mol} \times 18 \text{ g/mol} = 36 \text{ g} \][/tex]
So, the amount of water produced by the combustion of 16 grams of methane is 36 grams.
