Final answer:
To determine the amount of water vapor produced from burning 128 grams of methane, the stoichiometry of the balanced combustion reaction is used. Calculating the moles of methane and then the moles of water produced allows us to find the mass of water, which is approximately 288 grams.
Step-by-step explanation:
The student is asking about the products of a combustion reaction of methane with oxygen. When methane (CH4) is burned, it reacts with oxygen (O2) to produce carbon dioxide (CO2) and water vapor (H2O). To find out how many grams of water are produced when 128 grams of methane is burned, we can use the concept of stoichiometry. The balanced chemical reaction is as follows:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
First, we need to calculate the moles of methane in 128 grams:
Molar mass of CH4 = 12.01 (C) + 4 * 1.01 (H) = 16.05 g/mol
Moles of CH4 = 128 g / 16.05 g/mol ≈ 7.97 mol
According to the balanced equation, 1 mole of CH4 produces 2 moles of H2O. Thus, 7.97 moles of CH4 will produce 2 * 7.97 moles of H2O.
Total moles of H2O produced = 2 * 7.97 moles
Now, calculate the mass of water produced:
Molar mass of H2O = 2 * 1.01 (H) + 16.00 (O) = 18.02 g/mol
Mass of H2O = Total moles of H2O * Molar mass of H2O
= 2 * 7.97 moles * 18.02 g/mol ≈ 287.76 g
The answer is approximately 288 grams of water vapor, which corresponds with option D.