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 determine the maximum mass of water that can be converted to steam using the energy from the oxidation of methane, we first need the stoichiometry of methane combustion:

1. The balanced equation for methane combustion is:

   $CH_4 + 2 O_2 \rightarrow CO_2 + 2 H_2O$

   This tells us that one mole of methane ( ext{CH}_4) produces two moles of water ( ext{H}_2O).

2. We also know that the enthalpy change ( \Delta H) for the conversion of liquid water to steam is +44.0 kJ mol⁻¹. Thus for every mole of water converted to steam, 44.0 kJ of energy is required.

3. The amount of energy from the complete oxidation of one mole of methane is approximately 890 kJ.

4. Next, we can calculate the number of moles of water that can be converted to steam:

   $\text{Moles of water} = \frac{\text{Energy from methane}}{\Delta H} = \frac{890\text{ kJ}}{44.0\text{ kJ mol}^{-1}} \approx 20.23 \text{ moles}$

5. Now, since each mole of water has a mass of about 18 grams, we can find the maximum mass of the water:

   $\text{Mass of water} = 20.23\text{ moles} \times 18\text{ g mol}^{-1} \approx 364.14 \text{ g}$

Thus, rounding gives us a maximum mass of approximately 368 grams.

Therefore, the correct answer is D. 368.