Question

A volume of $\,200{\text{ }}ml\,$ oxygen is added to $\,100{\text{ }}ml\,$ of a mixture containing $\,C{S_2}\,$ vapour and $\,CO\,$, and the total mixture is burnt. After combustion, the volume of the entire mixture is $\,245{\text{ }}ml\,$. Calculate the volume of the oxygen that remains
A.$\,67.5ml\,$
B.$\,125.0ml\,$
C.$\,200.0ml\,$
D.$\,100.0ml\,$

Answer 1

Combustion involves complex series of radical elemental reactions. Fuel materials and oxygen must be available as well as an external source of energy to initiate the combustion cycle in order to proceed with the combustion reaction. While when combined with oxygen gas, certain content may spontaneously burst into flame, most substances need a spark or other source of energy to begin burning. The heat produced by the reaction is sufficient to proceed until the combustion reaction starts.

Complete step by step answer:
Let us assume total of $100ml$ mixture of $C{S_2}$ vapour and $CO$
In this taking “$\,100 - {\text{ }}X\,$ “$ml$ $C{S_2}$
“$\,X\,$” $ml$ of $CO$
First let us see the combustion reaction in $C{S_2}$
$$$$$C{S_2} + 3{O_2}\xrightarrow[{}]{}C{O_2} + 2S{O_2}$$,100 - X{\text{ }} + {\text{ }}3\left( {100 - X} \right)100 - X{\text{ }} + {\text{ }}2\left( {100 - X} \right),$Next writing combustion equation with$CO$$CO + \dfrac{1}{2}{O_2} \to C{O_2}$$,X + \dfrac{X}{2} \to X,$Total volume of${O_2}$is$,200ml,,$which is given in the question From both equation we are taking volume used by${O_2}$is “Y”$3(100 - X) + \dfrac{X}{2} = Y$(VOLUME USED BY${O_2}$) By simplifying we get$5X + 2Y = 600 \to (1)$Now volume${O_2}$which is not used = total volume used by${O_2}$-volume used by${O_2}$Volume of${O_2}$which is not used$, = 200 - Y,$After the combustion total volume of entire gas mixture which is given in question$, = 245ml,$From both combustion equation of$C{S_2}$and$CO$$(100 - X) + 2(100 - X) + X + (200 - Y) = 245$(Volume of entire gas mixture) By simplifying we get$2X + Y = 255 \to (2)$Multiplying equation (2) with 2 we get$4X + 2Y = 510 \to (3)$By solving equation (1) and (3) we get$X = 90ml$$,Y = 75ml,$Volume used by${O_2}$is$,Y = 75ml,$Volume of${O_2}$which is not used is$, = 200 - Y = 200 - 75 = 125ml,$Hence volume of${O_2}$which is not used or remains is$,125ml,$
Hence the correct answer is option B.

Additional information:
-$C{S_2}$ is highly flammable. Its combustion affords sulfur dioxide according to this ideal stoichiometry
Incomplete combustion occurs because of
-Insufficient mixing of air and fuel
-Insufficient air supply to the flame

Note: As substances spontaneously react with heat and light, the heat emitted fuels the mechanism to make the reaction violent and fast. It takes the form of a fire when little is done to regulate this process. It is this accidental and aggressive nature that separates combustion from most related processes that happens in the presence of oxygen.