Question

A gas mixture of ${\text{3litres}}$ of propane and butane on complete combustion at
{\text{2}}{{\text{5}}^{\text{^o }}}{\text{C}} produced ${\text{10 litres}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$. The gas mixture contains initially:
A. ${\text{2 litres }}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{3}}}{\text{ + 1litre }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$
B. ${\text{1litres }}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}{\text{ + 2litres }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$
C. ${\text{1}}{\text{.5litres }}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}{\text{ + 1}}{\text{.5litres }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$
D. ${\text{2}}{\text{.5litres }}{{\text{C}}_3}{{\text{H}}_8}{\text{ + 1}}{\text{.5litres }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$

Answer 1

As the formula of propane and butane is ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}$ and ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ respectively. As it is given that mixture of propane and butane is ${\text{3 litres}}$ and if we see the question after combustion ${\text{C}}{{\text{O}}_{\text{2}}}$ formed ${\text{10 litres}}$ .

Complete step by step answer: Here, if we make the equation of the given question
${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}{\text{ + }}{{\text{O}}_{\text{2}}} \to {\text{C}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
Balancing the above equation:
${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}{\text{ + 5}}{{\text{O}}_{\text{2}}} \to 3{\text{C}}{{\text{O}}_{\text{2}}}{\text{ + 4}}{{\text{H}}_{\text{2}}}{\text{O}}$
So, our second equation is
${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\text{ + }}{{\text{O}}_{\text{2}}} \to {\text{C}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
Balancing this equation also:
${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\text{ + }}\dfrac{{13}}{2}{{\text{O}}_{\text{2}}} \to 4{\text{C}}{{\text{O}}_{\text{2}}}{\text{ + 5}}{{\text{H}}_{\text{2}}}{\text{O}}$
Let suppose${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}{\text{ = x}}$.
So ${\text{1}}$ mole of propane produces ${\text{3}}$ moles of ${\text{C}}{{\text{O}}_{\text{2}}}$
Therefore, ${\text{x}}$ moles will produce ${\text{3x}}$ moles
${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\text{ = y}}$
So, ${\text{1}}$ mole of butane produces ${\text{4}}$ moles of ${\text{C}}{{\text{O}}_{\text{2}}}$
Therefore,${\text{y}}$ moles will produce ${\text{4 y}}$ moles.
Now if we make the equation of the question:
So, ${\text{x + y = 3}}$ ${\text{ - - - - - }}\left( {\text{1}} \right)$ (given)
And
Volume of ${\text{C}}{{\text{O}}_{\text{2}}}$formed ${\text{10 litres}}$
Therefore, ${\text{3x + 4y = 10 - - - - }}\left( {\text{2}} \right)$
So by solving $\left( {\text{1}} \right){\text{and}}\left( {\text{2}} \right)$ equations
From $\left( {\text{1}} \right)$ equation, we get ${\text{x + y = 3}}$
${\text{x = 3 - y}}$ ${\text{ - - - - *}}$
From $\left( {\text{2}} \right)$ equation, ${\text{3x + 4y = 10}}$
Putting the value of ${\text{'x'}}$ in above equation
${\text{3}}\left( {{\text{3 - y}}} \right){\text{ + 4y = 10}}$
Now solving above equation
${\text{9 - 3y + 4y = 10}}$
$\boxed{{\text{y = 1}}}$
Now putting the value of ${\text{'y'}}$ in ${\text{ - - - - *}}$
${\text{x = 3 - y}}$
${\text{x = 3 - 1}}$
$\boxed{{\text{x = 2}}}$
By solving the equation, we get ${\text{x = 2L and y = 1L}}$
So, the composition of propane in the gas mixture is ${\text{2L}}$ and butane is${\text{1L}}$.
So, the correct answer is “Option A”.

Note: volume is directly proportional to number of moles when temperature and pressure are constant.