Question

10ml of a gaseous hydrocarbon on combustion gives 40ml of $ {\text{C}}{{\text{O}}_2} $ and 50ml $ {{\text{H}}_2}{\text{O}} $ vapour STP. The hydrocarbon is;
A) $ {{\text{C}}_2}{{\text{H}}_6} $
B) $ {{\text{C}}_8}{{\text{H}}_{10}} $
C) $ {{\text{C}}_4}{{\text{H}}_8} $
D) $ {{\text{C}}_4}{{\text{H}}_{10}} $

Answer 1

Hint :Combustion is a chemical reaction in which a substance rapidly reacts with oxygen, the oxygen is mainly sourced from the air or any oxidizer and forms heat. A hydrocarbon is made up of hydrogen and carbon(as constituents), writing the balanced reaction of the given combustion reaction to reach the answer.

Complete Step By Step Answer:
Since we do not know what is the formula of the hydrocarbon, so we will assume a general for it,
Let us consider that the formula of the hydrocarbon is $ {{\text{C}}_{\text{X}}}{{\text{H}}_{\text{Y}}} $ , where x and y are the value we need to find out.
As mentioned in hint now we will write the balanced chemical reaction of combustion of this hydrogen;
 $ {{\text{C}}_{\text{X}}}{{\text{H}}_{\text{Y}}} + {\text{ }}\left( {{\text{x + }}\dfrac{{\text{y}}}{2}} \right){{\text{O}}_2} \to {\text{xC}}{{\text{O}}_2}{\text{ + }}\dfrac{{\text{y}}}{2}{{\text{H}}_2}{\text{O}} $
(‘x’ is the number of carbon present in the reactant side therefore it should also be present on the product side and ‘y’ is the number of hydrogen present on the reactant side, but on the product side 2 hydrogens are present, therefore to balance out the hydrogen on both side we have written $ \dfrac{{\text{y}}}{2} $ in the product side)
Now, according to the given conditions, we can write the following,
 $
{\text{ }}{{\text{C}}_{\text{X}}}{{\text{H}}_{\text{Y}}} + {\text{ }}\left( {{\text{x + }}\dfrac{{\text{y}}}{2}} \right){{\text{O}}_2} \to {\text{xC}}{{\text{O}}_2}{\text{ + }}\dfrac{{\text{y}}}{2}{{\text{H}}_2}{\text{O}} \\
{\text{(at, t = 0) 10 0 0}} \\
({\text{at, t = }}\infty ){\text{ 0 x10 }}\dfrac{{\text{y}}}{2}10 \\
 $
where t = 0 (time at initial stage)
and t = $ \infty $ (time at which the reactant has completely being converted to product)
 $ \Rightarrow $ 10x = 40 (this statement means that the carbon dioxide formed in the final stage is equal to the 40ml formed given in our question)
 $ \Rightarrow $ x = $ \dfrac{{40}}{{10}} $
 $ \Rightarrow $ x = 4
Similarly, we can calculate for water
 $ \Rightarrow $ $ \dfrac{{\text{y}}}{2}10 $ = 50
 $ \Rightarrow $ 10y = 100
 $ \Rightarrow $ y = 10
Therefore, form the above calculation we can write the formula of the hydrocarbon $ $
as $ {{\text{C}}_4}{{\text{H}}_{10}} $ by this our chemical reaction becomes;
 $ {{\text{C}}_4}{{\text{H}}_{10}} + {\text{ }}\left( {{\text{ }}\dfrac{{13}}{2}} \right){{\text{O}}_2} \to 4{\text{C}}{{\text{O}}_2}{\text{ + 5}}{{\text{H}}_2}{\text{O}} $
Hence, the correct answer is option (D)i.e., $ {{\text{C}}_4}{{\text{H}}_{10}} $

Note :
Combustion is an irreversible reaction, hence we know at some point in time the reactant would completely be changed into the products. The condition is not similar for reversible reaction, because there will be always some reactant present in the solution mixture.