<u>Answer:</u> The mass of methane burned is 12.4 grams.
<u>Explanation:</u>
The chemical equation for the combustion of methane follows:
![CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)](https://tex.z-dn.net/?f=CH_4%28g%29%2B2O_2%28g%29%5Crightarrow%20CO_2%28g%29%2B2H_2O%28g%29)
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:
![\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_f_{(CH_4(g))}=-74.81kJ/mol\\\Delta H^o_f_{O_2}=0kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_f_%7B%28H_2O%28g%29%29%7D%3D-241.82kJ%2Fmol%5C%5C%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%3D-393.51kJ%2Fmol%5C%5C%5CDelta%20H%5Eo_f_%7B%28CH_4%28g%29%29%7D%3D-74.81kJ%2Fmol%5C%5C%5CDelta%20H%5Eo_f_%7BO_2%7D%3D0kJ%2Fmol)
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:
![(1):H_2O(l)(26^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(3):H_2O(g)(100^oC)\rightarrow H_2O(g)(101^oC)](https://tex.z-dn.net/?f=%281%29%3AH_2O%28l%29%2826%5EoC%29%5Crightarrow%20H_2O%28l%29%28100%5EoC%29%5C%5C%5C%5C%282%29%3AH_2O%28l%29%28100%5EoC%29%5Crightarrow%20H_2O%28g%29%28100%5EoC%29%5C%5C%5C%5C%283%29%3AH_2O%28g%29%28100%5EoC%29%5Crightarrow%20H_2O%28g%29%28101%5EoC%29)
Now, we calculate the amount of heat released or absorbed in all the processes.
![q_1=mC_p,l\times (T_2-T_1)](https://tex.z-dn.net/?f=q_1%3DmC_p%2Cl%5Ctimes%20%28T_2-T_1%29)
where,
= amount of heat absorbed = ?
m = mass of water = 242.0 g
= specific heat of water = 4.18 J/g°C
= final temperature = ![100^oC](https://tex.z-dn.net/?f=100%5EoC)
= initial temperature = ![26^oC](https://tex.z-dn.net/?f=26%5EoC)
Putting all the values in above equation, we get:
![q_1=242.0g\times 4.18J/g^oC\times (100-(26))^oC=74855.44J](https://tex.z-dn.net/?f=q_1%3D242.0g%5Ctimes%204.18J%2Fg%5EoC%5Ctimes%20%28100-%2826%29%29%5EoC%3D74855.44J)
![q_2=m\times L_v](https://tex.z-dn.net/?f=q_2%3Dm%5Ctimes%20L_v)
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:
![q_2=242g\times 2257J/g=546194J](https://tex.z-dn.net/?f=q_2%3D242g%5Ctimes%202257J%2Fg%3D546194J)
![q_3=mC_p,g\times (T_2-T_1)](https://tex.z-dn.net/?f=q_3%3DmC_p%2Cg%5Ctimes%20%28T_2-T_1%29)
where,
= amount of heat absorbed = ?
m = mass of steam = 242.0 g
= specific heat of steam = 2.08 J/g°C
= final temperature = ![101^oC](https://tex.z-dn.net/?f=101%5EoC)
= initial temperature = ![100^oC](https://tex.z-dn.net/?f=100%5EoC)
Putting all the values in above equation, we get:
![q_3=242.0g\times 2.08J/g^oC\times (101-(100))^oC=503.36J](https://tex.z-dn.net/?f=q_3%3D242.0g%5Ctimes%202.08J%2Fg%5EoC%5Ctimes%20%28101-%28100%29%29%5EoC%3D503.36J)
Total heat required = ![q_1+q_2+q_3=(74855.44+546194+503.36)=621552.8J=621.552kJ](https://tex.z-dn.net/?f=q_1%2Bq_2%2Bq_3%3D%2874855.44%2B546194%2B503.36%29%3D621552.8J%3D621.552kJ)
- 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 = ![\frac{1}{802.34}\times 621.552=0.775mol](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B802.34%7D%5Ctimes%20621.552%3D0.775mol)
To calculate the number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Molar mass of methane = 16 g/mol
Moles of methane = 0.775 moles
Putting values in above equation, we get:
![0.775mol=\frac{\text{Mass of methane}}{16g/mol}\\\\\text{Mass of methane}=(0.775mol\times 16g/mol)=12.4g](https://tex.z-dn.net/?f=0.775mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20methane%7D%7D%7B16g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20methane%7D%3D%280.775mol%5Ctimes%2016g%2Fmol%29%3D12.4g)
Hence, the mass of methane burned is 12.4 grams.