Question

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 1

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.