Final answer:
The mass of water produced from the combustion of 8.00x10⁻³ g of methane is calculated using the balanced thermochemical equation for methane combustion. The molar mass of methane is used to convert the given mass to moles, which is then used to find the moles of water produced, ultimately converted to grams.
Step-by-step explanation:
The mass of water produced from the complete combustion of 8.00x10⁻³ g of methane ([CH₄]) can be calculated using the balanced thermochemical equation for the combustion of methane:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 890.4 kJ
This equation tells us that 1 mole of methane ([CH₄]), which has a mass of approximately 16.04 g/mol, produces 2 moles of water ([H₂O]). Therefore, to find out how much water is produced from 8.00x10⁻³ g of methane, we first calculate the number of moles of methane:
number of moles of CH₄ = mass of CH₄ (g) / molar mass of CH₄ (g/mol)
number of moles of CH₄ = 8.00x10⁻³ g / 16.04 g/mol
Since the ratio of methane to water in the balanced equation is 1:2, we then multiply the number of moles of methane by 2 to get the number of moles of water produced, and convert this to grams using the molar mass of water (approximately 18.02 g/mol).
mass of water (H₂O) = number of moles of water × molar mass of H₂O