Complete combustion of 25.0 L (30 C, 100.8 kPa) of a methane-ethane mixture (40% ethane by mass) in oxygen produces 1055 kJ of heat. Combustion of an equal amount of methane alone produces only 882 kJ of heat. Estimate the specific heat of combustion of higher alkanes (in kJ/mol).
As I have the answer and I did it similarly I will just copy the answer:
There was 1 mole of the mixture, n(ethane)=0.262 mol and n(methane)=0.738 mol.
From the heat released: 882·0.738 + х·0.262 = 1055, x=1542 kJ/mol (enthalpy of combustion of ethane).
Comparing the structure of ethane and methane we see that the heat of combustion of the CH2 unit is 1542-882=660 kJ/mol. The general formula of alkanes is CnH2n+2 or, equivalently, [CH4 + (n-1)CH2]. Thus, 1 mol of CnH2n+2 will produce 882 + (n-1)*660 = 222 + 660*n kJ of heat...
Until here everything is ok, but now I am confused:
...The molar mass of CnH2n+2 is (14n+2) g/mol. Combustion of 1 kg of an alkane produces (222+660n)/(14n+2) kJ of heat. As n increases indefinitely, the value of the fraction approaches 660n/14n=47. Thus, the specific heat of combustion of higher alkanes is 47 MJ/mol.
Isn't 222 + 660*n kJ/mol the answer? Why doing the extra calculations? Can someone explain why is 1kg used?