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The heat combustion of butane is \[2880kJ{\text{ }}mo{l^{ - 1}}\]. What is the heat liberated by burning \[1kg\] of butane in excess of oxygen supply?
A. \[49655{\text{ }}kJ\]
B. \[4965{\text{ }}kJ\]
C. \[4655{\text{ }}kJ\]
D. \[9655{\text{ }}kJ\]

Answer
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Hint: The heat of combustion is defined as the energy which is liberated when a substance undergoes complete combustion, complete combustion here refers to an excess amount of oxygen and the temperature is standard that is \[{25^o}C\].

Complete answer:
In the question it’s given that the heat of combustion of butane is \[2880kJ{\text{ }}mo{l^{ - 1}}\]. So initially we will calculate the molecular mass of butane which can be calculated by adding the molecular mass according to the chemical formula, the chemical formula for butane is \[{C_4}{H_{10}}\] , Hence the molecular mass is:
\[12 \times 4 + 1 \times 10 = 58g\]
The molecular mass of one carbon atom is $12g$ and of one hydrogen atom is $1g$ .
Hence for 58g of butane our heat of combustion is \[2880kJ{\text{ }}mo{l^{ - 1}}\]. Now, we are asked the heat of combustion for \[1kg\] of butane that means \[1000{\text{ }}g\] of butane. Hence, we will simply apply the unitary method and we will calculate the heat of combustion for $1g$ of butane followed by the heat of combustion of \[1000{\text{ }}g\] of butane.
As for \[58{\text{ }}g\] of butane heat of combustion is \[2880kJ{\text{ }}mo{l^{ - 1}}\]
Then for $1g$ of butane heat of combustion will be \[\dfrac{{2880}}{{58}}\]
Similarly, for \[1000{\text{ }}g\] of butane heat of combustion will be \[\dfrac{{2880}}{{58}} \times 1000 = 49655{\text{ }}KJ\]
So, the correct value of heat of combustion is \[49655{\text{ }}kJ\], Option A.

Note:
While using the unitary method a student has to keep in mind the value that one needs to calculate should be at the right-hand side for an easy calculation. As here we needed to calculate the heat of combustion and the given values were the masses, Hence, we placed masses at the left-hand side and Heat of combustion at the right-hand side.