Question

$20{\text{mL}}$ of methane is completely burnt using $50\,{\text{mL}}$ of oxygen. The volume of the gas left after cooling to room temperature is:
A. ${\text{80}}\,{\text{ml}}$
B. ${\text{40 ml}}$
C. ${\text{60 ml}}$
D. ${\text{30ml}}$

Answer 1

Hint: To determine the volume of the gas left balanced equation is required. After writing the balanced equation, by comparing the volume of reactant and product, the amount of gas left can be determined.

Complete step by step answer:
$20{\text{mL}}$ of methane is completely burnt using $50\,{\text{mL}}$ of oxygen at room temperature.
The reaction between methane and oxygen is as follows:
${\text{C}}{{\text{H}}_{\text{4}}}{\text{(g)}}\,{\text{ + }}\,2{{\text{O}}_2}{\text{(g)}}\,\, \to {\text{C}}{{\text{O}}_2}{\text{(g)}}\, + \,2\,{{\text{H}}_2}{\text{O(l)}}$
According to the balanced reaction, one mole of methane is reacting with two moles of oxygen gas and forms one mole of carbon dioxide gas and two moles of water.
According to the ideal gas equation, the volume is directly proportional to the number of moles.
So, we can use volume in place of the number of moles.
So, according to the balanced reaction, one volume of methane is reacting with two volumes of oxygen gas and forms one volume of carbon dioxide gas and two volumes of water.
So, determine the volume of oxygen needed to react with ${\text{20}}\,{\text{ml}}$ methane as follows:
\[{\text{1V}}\,\,{\text{C}}{{\text{H}}_{\text{4}}}{\text{(g)}}\,\, \to 2{\text{V}}\,{{\text{O}}_2}{\text{(g)}}\,\]
\[\Rightarrow {\text{20}}\,{\text{ml}}\,\,{\text{C}}{{\text{H}}_{\text{4}}}{\text{(g)}}\,\, \to 40\,{\text{ml}}\,{{\text{O}}_2}{\text{(g)}}\,\]
So, ${\text{20}}\,{\text{ml}}$ methane will react with ${\text{40}}\,{\text{ml}}$ oxygen gas.
So, the left amount of oxygen gas is,
The given amount of the oxygen gas is ${\text{50}}\,{\text{ml}}$.
${\text{50}}\,{\text{ml}}\,{\text{ - 40}}\,{\text{ml}}\,{\text{ = }}\,{\text{10 ml}}$
So, the left volume of oxygen gas is $\,{\text{10 ml}}$.
Determine the volume of carbon dioxide gas formed by ${\text{20}}\,{\text{ml}}$ methane as follows:
\[{\text{1V}}\,\,{\text{C}}{{\text{H}}_{\text{4}}}{\text{(g)}}\,\, \to 1{\text{V}}\,{\text{C}}{{\text{O}}_2}{\text{(g)}}\,\]
\[\Rightarrow {\text{20}}\,{\text{ml}}\,\,{\text{C}}{{\text{H}}_{\text{4}}}{\text{(g)}}\,\, \to 20\,{\text{ml}}\,{\text{C}}{{\text{O}}_2}{\text{(g)}}\,\]
So, \[20\,{\text{ml}}\] carbon dioxide forms by the reaction of ${\text{20}}\,{\text{ml}}$ methane.
After completion of the reaction, the gases remaining in the reaction mixture are carbon dioxide and the left amount of oxygen gas.
So, the total volume of the gas is,
\[20\,{\text{ml}}\,{\text{C}}{{\text{O}}_2}{\text{(g)}}\,{\text{ + }}\,10\,{\text{ml}}\,{{\text{O}}_2}{\text{(g)}}\,{\text{ = }}\,30\,{\text{ml}}\,{\text{g}}\].
The volume of the gas left after cooling to room temperature is \[30\,{\text{ml}}\].

Therefore, option (D) ${\text{30ml}}$ is correct.

Note:
The ideal gas equation is as follows: $PV = nRT$. The pressure and temperature are the same for all the rating species, so the volume is directly proportional to the number of moles. To compare the moles or volume, the balanced equation is necessary. The burning of methane in presence of oxygen gas is known as decomposition. These reactions are known as a decomposition reaction. The decomposition of hydrocarbons in presence of oxygen gives carbon dioxide and water.