Question

Given,\[10ml\] of a mixture of \[CO,C{H_4},\] and \[{N_2}\] exploded with an excess of oxygen gas gave a contraction of \[6.5ml\] . There was a further contraction of \[7ml\] , when the residual gas treated with \[KOH\] . Find the volume of \[CO,C{H_4},\] and \[{N_2}\] respectively is:

Answer 1

Hint: Given that \[10ml\] of a mixture of \[CO,C{H_4},\] and \[{N_2}\] exploded with an excess of oxygen gas gave a contraction of \[6.5ml\] . This means carbon monoxide treated with oxygen gas produces carbon dioxide. The moles can be determined from this equation. Further contraction means methane reacts with oxygen gas and produces carbon dioxide and water. From this the equation of volume can be determined. From these two equations the volume of gases can be determined.

Complete answer:
Combustion reaction means the addition of alkanes treated with oxygen gas produces carbon dioxide and water vapor.
\[CO + \dfrac{1}{2}{O_2} \to C{O_2}\]
The above equation represents \[x\] moles of carbon monoxide and \[\dfrac{x}{2}\] moles of oxygen were reacted to produce \[x\] moles of carbon dioxide.
\[C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\]
The above equation represents \[y\] moles of methane, and \[2y\] moles of oxygen gas were reacted to produce \[y\] moles of carbon dioxide.
Given that \[10ml\] of a mixture of \[CO,C{H_4},\] and \[{N_2}\] exploded with an excess of oxygen gas gave a contraction of \[6.5ml\] .
It can be written as \[\dfrac{x}{2} + 2y = 6.5\]
Further contraction at \[10ml\] means
\[x + y = 7\]
By solving the above two equations,
\[x = 5ml;y = 2ml\]
Volume of \[CO\] is \[5ml\]
Volume of \[C{H_4}\] is \[2ml\]
Given that the total mixture is \[10ml\]
Thus, volume of \[{N_2}\] is \[10 - 7 = 3ml\]

Note:
At first, only carbon monoxide reacts with oxygen gas to produce carbon dioxide and then further methane gas is treated with oxygen gas to produce carbon dioxide and water. But nitrogen gas does not react with oxygen gas as it requires a very high temperature. Thus, nitrogen is an unreactive gas.