<u>Answer:</u> The mass of methane burned is 12.4 grams.
<u>Explanation:</u>
The chemical equation for the combustion of methane follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CO_2(g))})+(2\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(CH_4(g))})+(2\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CH_4%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-393.51))+(2\times (-241.82))]-[(1\times (-74.81))+(2\times (0))]\\\\\Delta H^o_{rxn}=-802.34kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-393.51%29%29%2B%282%5Ctimes%20%28-241.82%29%29%5D-%5B%281%5Ctimes%20%28-74.81%29%29%2B%282%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-802.34kJ)
The heat calculated above is the heat released for 1 mole of methane.
The process involved in this problem are:

Now, we calculate the amount of heat released or absorbed in all the processes.

where,
= amount of heat absorbed = ?
m = mass of water = 242.0 g
= specific heat of water = 4.18 J/g°C
= final temperature = 
= initial temperature = 
Putting all the values in above equation, we get:


where,
= amount of heat absorbed = ?
m = mass of water or steam = 242 g
= latent heat of vaporization = 2257 J/g
Putting all the values in above equation, we get:


where,
= amount of heat absorbed = ?
m = mass of steam = 242.0 g
= specific heat of steam = 2.08 J/g°C
= final temperature = 
= initial temperature = 
Putting all the values in above equation, we get:

Total heat required = 
- To calculate the number of moles of methane, we apply unitary method:
When 802.34 kJ of heat is needed, the amount of methane combusted is 1 mole
So, when 621.552 kJ of heat is needed, the amount of methane combusted will be = 
To calculate the number of moles, we use the equation:

Molar mass of methane = 16 g/mol
Moles of methane = 0.775 moles
Putting values in above equation, we get:

Hence, the mass of methane burned is 12.4 grams.