In a gas-fired power station, the energy available from the above reaction is used to convert water in a boiler from liquid water to steam, H₂O(l) → H₂O(g) ΔH = +44.0 kJ mol⁻¹ The maximum mass of water, in grams, that could be converted from liquid water to steam by the complete oxidation of one mole of methane is A - VCE - SSCE Chemistry - Question 4 - 2004 - Paper 1

To solve this problem, we need to determine how much heat is produced by the complete oxidation of one mole of methane (CH₄) and then use that to find out how much water can be converted to steam.

1. Determine the Heat Released by Methane: The balanced chemical equation for the complete combustion of methane is:

   $CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)$

   The combustion of one mole of methane releases approximately 890 kJ of energy.

2. Calculate the Energy Needed to Convert Water to Steam: The enthalpy change for converting liquid water to steam at 100°C is given as +44.0 kJ/mol. This is the amount of energy required to convert one mole of water to steam.

3. Calculate How Many Moles of Water Can Be Vaporized: Now we calculate how many moles of water can be vaporized using the total energy from the combustion of methane:

   Let the number of moles of water vaporized be n:

   n \\ = 20.22727 ext{ mol}$$

4. Convert Moles of Water to Grams: To find the mass in grams, multiply the number of moles by the molar mass of water:

   Molar mass of H₂O = 18 g/mol.

   $Mass = n imes ext{Molar mass} = 20.22727 ext{ mol} imes 18 ext{ g/mol} = 364.0909 ext{ g}$

5. Choose the Closest Answer: Rounding down to match the provided options, we see that 368 g is the maximum mass of water that can be vaporized, which corresponds to option D.