Question

A gas mixture of $3{\text{ }}liters$ of propane and butane on complete combustion at ${25^0}C$ produce $10{\text{ }}liters$ of $C{O_2}$ initial composition of the propane and butane in the gas mixture is:
A. $66.67\% ,{\text{ }}33.33\% \;$
B. $33.33\% ,{\text{ }}66.67\%$
C. $50\% ,{\text{ }}50\%$
D. $60\% ,{\text{ }}40\%$

Answer 1

Combustion reaction to the type of reaction that causes a flame. Combustion is a chemical reaction between a fuel and an oxidant, usually atmospheric oxygen, high-temperature exothermic (heat releasing) redox (oxygen adding) which creates oxidized, often gaseous products in a mixture called smoke.

Complete answer:
As per given gas mixture of $3{\text{ }}liters$ of propane and butane on complete combustion, we know the formula of propane and Butane is ${C_3}{H_8}{\text{ }}and{\text{ }}{C_4}{H_{10}}$ respectively. Now the question is after combustion of $C{O_2}$ formed $10{\text{ }}litres.$
First, we can make the reaction -
The reactions of propane with oxygen: ${C_3}{H_{8\;}} + {\text{ }}{O_{2\;{\text{ }}}} - \;C{O_2}\; + {\text{ }}{H_2}O\;\;\;\;\;\;\;\;\;\;$
Balancing the equation ${C_3}{H_{8\;}} + {\text{ }}5{O_{2\;}}\; \to \;3C{O_{2\;}} + 4{H_2}O\; - - - - - - - - - \left( 1 \right)$
the reactions of butane with oxygen: ${C_4}{H_{10}}\; + {\text{ }}{O_2}\;\;\; \to \;\;{H_2}O{\text{ }} + \;C{O_2}$
Balancing the equation$\;{C_4}{H_{10}}\; + {\text{ }}\left( {\dfrac{{13}}{2}} \right){\text{ }}{O_2} \to 5{H_2}O{\text{ }} + {\text{ }}4C{O_{2\;}}\;\;\;\; - - - - - - - - - \left( 2 \right)$
Let suppose ${C_3}{H_8} = X\, moles$ and ${C_4}{H_{10}} = Y\, moles$
As per reaction of propane ${C_3}{H_{8\;}} + {\text{ }}5{O_{2\;}}\; \to \;3C{O_{2\;}} + 4{H_2}O\; - - - - - - - - - \left( 1 \right)$
so one mole of propane produces =$\;3$ moles of$C{O_2}$ .
$\therefore X$ Moles of propane produce = $3X$ moles of $C{O_2}$
As per reaction of butane $\;{C_4}{H_{10}}\; + {\text{ }}\left( {\dfrac{{13}}{2}} \right){\text{ }}{O_2} \to 5{H_2}O{\text{ }} + {\text{ }}4C{O_{2\;}}\;\;\;\; - - - - - - - - - \left( 2 \right)$
So one mole of Butane produces =$4$ moles of$C{O_2}$
$\therefore Y\;$ moles of Butane produce = $4Y$ moles of $C{O_2}$
Now $X + Y{\text{ }} = 3\;\; - - - - - - - \left( 3 \right)$ Given,
So, $3X{\text{ }} + 4Y{\text{ }} = {\text{ }}10\; - - - - - - - \left( 4 \right)$ $\;\therefore$ Given Volume of $C{O_2}\; = 10{\text{ }}litres\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$
Solving equation (3) and (4)
$X + Y{\text{ }} = 3\;\; - - - - - - - \left( 3 \right)$
$\Rightarrow X = {\text{ }}3 - Y$
From equation $3X{\text{ }} + 4Y{\text{ }} = {\text{ }}10\; - - - - - - - \left( 4 \right)$
Putting the value of $X$ in equation$\left( 4 \right)$
$\Rightarrow 3\left( {3 - Y} \right) + 4Y = {\text{ }}10$
By solving, $Y = 1,$ now putting the value of $Y$ in equation $\left( 3 \right)$ we get, $X = {\text{ }}2$
So, the composition of propane in gas mixture is $2\,L$ and butane is $1\,L.$
$\therefore$Volume of propane = $2\,L.$
$\therefore$Volume of butane = $1\,L.$
Percentage of propane = $\dfrac{{Volume{\text{ }}of{\text{ }}propane}}{{total{\text{ }}gas{\text{ }}mixture{\text{ }}volume}} \times 100$ = $\dfrac{2}{{1 + 2}} \times 100 = 66.67\%$
Percentage of butane =${\text{ }}\dfrac{{Volume{\text{ }}of{\text{ }}butane}}{{total{\text{ }}gas{\text{ }}mixture{\text{ }}volume}} \times 100$ = $\dfrac{1}{{1 + 2}} \times 100 = 33.33\%$

**So the option (A) is correct

Note:**
Combustion reaction is an exothermic reaction in the form of heat and light that releases energy. This releases the full amount of energy from the fuel being reacted when a gas undergoes complete combustion.