<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:
We are given:
Putting values in above equation, we get:
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.
- <u>For process 1:</u>
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:
- <u>For process 2:</u>
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:
- <u>For process 3:</u>
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.