Respuesta :
Answer: The mass of methane burned is 12.4 grams.
Explanation:
The chemical equation for the combustion of methane follows:
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\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))})][/tex]
We are given:
[tex]\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[/tex]
Putting values in above equation, we get:
[tex]\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[/tex]
The heat calculated above is the heat released for 1 mole of methane.
The process involved in this problem are:
[tex](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)[/tex]
Now, we calculate the amount of heat released or absorbed in all the processes.
- For process 1:
[tex]q_1=mC_p,l\times (T_2-T_1)[/tex]
where,
[tex]q_1[/tex] = amount of heat absorbed = ?
m = mass of water = 242.0 g
[tex]C_{p,l}[/tex] = specific heat of water = 4.18 J/g°C
[tex]T_2[/tex] = final temperature = [tex]100^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]26^oC[/tex]
Putting all the values in above equation, we get:
[tex]q_1=242.0g\times 4.18J/g^oC\times (100-(26))^oC=74855.44J[/tex]
- For process 2:
[tex]q_2=m\times L_v[/tex]
where,
[tex]q_2[/tex] = amount of heat absorbed = ?
m = mass of water or steam = 242 g
[tex]L_v[/tex] = latent heat of vaporization = 2257 J/g
Putting all the values in above equation, we get:
[tex]q_2=242g\times 2257J/g=546194J[/tex]
- For process 3:
[tex]q_3=mC_p,g\times (T_2-T_1)[/tex]
where,
[tex]q_3[/tex] = amount of heat absorbed = ?
m = mass of steam = 242.0 g
[tex]C_{p,g}[/tex] = specific heat of steam = 2.08 J/g°C
[tex]T_2[/tex] = final temperature = [tex]101^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]100^oC[/tex]
Putting all the values in above equation, we get:
[tex]q_3=242.0g\times 2.08J/g^oC\times (101-(100))^oC=503.36J[/tex]
Total heat required = [tex]q_1+q_2+q_3=(74855.44+546194+503.36)=621552.8J=621.552kJ[/tex]
- 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 = [tex]\frac{1}{802.34}\times 621.552=0.775mol[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of methane = 16 g/mol
Moles of methane = 0.775 moles
Putting values in above equation, we get:
[tex]0.775mol=\frac{\text{Mass of methane}}{16g/mol}\\\\\text{Mass of methane}=(0.775mol\times 16g/mol)=12.4g[/tex]
Hence, the mass of methane burned is 12.4 grams.
