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

Hint: 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.