Question

A cylinder of gas supplied by a company contains $14kg$ of butane. The heat of combustion of butane is $2658kJmo{l^{ - 1}}$ . A normal family requires $20MJ$ of energy per day for cooking. If the cylinder lasts for 26 days, the percentage of gas that is wasted due to insufficient combustion is _______.

Answer 1

Hint: We have to calculate the moles of butane from grams of butane using the molar mass of butane. We have to calculate the maximum energy which could be obtained using the heat of combustion and moles of butane. Next, we should calculate energy consumed by the family using the energy per day and total number of days. We have to calculate the percentage of gas wasted using the maximum energy which could be obtained and energy consumed by the family multiplied by 100.

Complete answer:
Given data contains,
Mass of butane is $14kg$.
The heat of combustion of butane is $2658kJmo{l^{ - 1}}$.
Energy required per day for cooking is $20MJ$.
Number of days is twenty six.
We know that the molar mass of butane is $58.12g/mol$.
We can now calculate the moles of butane using the molar mass and grams of butane.
We can write the formula to calculate the moles of butane.
$Moles = \dfrac{{Grams}}{{Molarmass}}$
Let us now substitute the values of grams of butane in the expression to calculate the moles.
$ \Rightarrow Moles = \dfrac{{14000g}}{{58.12g/mol}}$
On simplifying we get,
$ \Rightarrow Moles = 240.88mol$
The moles of butane is $240.88mol$.
Let us now calculate the maximum energy which could be obtained when there is no wastage using the heat of combustion and moles of butane.
The maximum energy which could be obtained= $2658\dfrac{{kJ}}{{mol}} \times 240.88mol$
The maximum energy which could be obtained= $640259kJ$
We have calculated the maximum energy which could be obtained when there is no wastage as $640259kJ$.
We have to convert the value in kilojoules to megajoules by multiplying the value with 1000.
$MJ = 640259kJ \times \dfrac{{1000MJ}}{{1kJ}}$
On simplifying we get,
$ \Rightarrow MJ = 640.259MJ$
We have calculated the maximum energy which could be obtained when there is no wastage as $640.259MJ$.
Let us now calculate the energy used by the family using the energy required for cooking per day and the number of days.
Energy used by the family= $20MJ/day \times 26days$
Energy used by the family= $520MJ$
The energy used by the family is $520MJ$.
So, the percentage of gas wasted because of insufficient combustion is calculated as,
Percentage of gas= $\dfrac{{640.259 - 520}}{{640.259}} \times 100$
On simplifying we get,
Percentage of gas= $18.78\% $
The percentage of gas wasted because of insufficient combustion is $18.78\% $.

Note:
One must remember to convert the kilojoules to megajoules using the conversion factor. Generally, combustion is an exothermic redox reaction that happens between oxidant and fuel. The oxidant is generally atmospheric oxygen which oxidizes and the products obtained are generally in gaseous phase.