Final answer:
The mass of water produced from the complete combustion of 5.10×10⁻³ g of methane is calculated by first determining the moles of methane, then using the stoichiometry of the balanced equation to find the moles of water, and finally converting this to mass using water's molar mass.
Step-by-step explanation:
To determine the mass of water produced from the complete combustion of 5.10×10⁻³ g of methane, we start with the balanced chemical equation for the combustion of methane:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
From the equation, one mole of methane produces two moles of water. First, we must find the number of moles of methane that corresponds to 5.10×10⁻³ g. We use the molar mass of methane (16.04 g/mol) in this calculation:
• Moles of methane = mass (g) / molar mass (g/mol) = 5.10×10⁻³ g / 16.04 g/mol
Next, knowing that two moles of water are produced for each mole of methane, we multiply the number of moles of methane by two to find the moles of water produced:
• Moles of water = Moles of methane × 2
Finally, we convert the moles of water to mass using the molar mass of water (18.02 g/mol):
• Mass of water (g) = Moles of water × molar mass of water (g/mol)
After calculating, you will get a value in grams that represents the mass of water produced from the combustion of 5.10×10⁻³ g of methane, which will be much less than any of the provided answer options A through D.