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Question: When ethane is burnt in excess of air, the products of combustion are: (a)- \(C\) and \({{H}_{2}}O...

When ethane is burnt in excess of air, the products of combustion are:
(a)- CC and H2O{{H}_{2}}O
(b)- COCO and H2O{{H}_{2}}O
(c)- CO2C{{O}_{2}} and H2{{H}_{2}}
(d)-CO2C{{O}_{2}} and H2O{{H}_{2}}O

Explanation

Solution

The general reaction of complete combustion of alkane is CnH2n+2+(3n+12)O2nCO2+(n+1)H2O{{C}_{n}}{{H}_{2n+2}}+\left( \dfrac{3n+1}{2} \right){{O}_{2}}\to nC{{O}_{2}}+(n+1){{H}_{2}}O. In ethane, there are 2 carbon atoms.

Complete step by step answer:
Ethane is a member of the alkane group and it has 2 carbon atoms.
On heating, alkanes readily burn in air or oxygen producing carbon dioxide (CO2C{{O}_{2}}) and water (H2O{{H}_{2}}O). This process is called combustion.
For example, Methane of combustion forms carbon dioxide and 2 moles of water. The reaction is:
CH4+O2CO2+2H2OC{{H}_{4}}+{{O}_{2}}\to C{{O}_{2}}+2{{H}_{2}}O
The general reaction of complete oxidation of alkane is:
CnH2n+2+(3n+12)O2nCO2+(n+1)H2O{{C}_{n}}{{H}_{2n+2}}+\left( \dfrac{3n+1}{2} \right){{O}_{2}}\to nC{{O}_{2}}+(n+1){{H}_{2}}O.
So, in ethane, the n is 2. So, the reaction is:
C2H6+72O22CO2+3H2O{{C}_{2}}{{H}_{6}}+\dfrac{7}{2}{{O}_{2}}\to 2C{{O}_{2}}+3{{H}_{2}}O.
So, the products of complete oxidation of ethane are 2 moles of carbon dioxide, and 3 moles of water.

Hence, the correct answer is an option (d)- CO2C{{O}_{2}} andH2O{{H}_{2}}O.

Additional information: There are 2 more forms of oxidation of alkane:
(i)- Incomplete combustion: (a)- If the combustion of alkanes is carried out in a limited supply of air, carbon monoxide is formed with unburnt carbon in the form of carbon black or soot.
2CH4+3O22CO+4H2O2C{{H}_{4}}+3{{O}_{2}}\to 2CO+4{{H}_{2}}O.
(b)- When the alkanes react with steam at 1273 K in the presence of nickel as a catalyst forming a mixture of carbon monoxide and hydrogen gas, is called syngas.
CH4+H2O1273K,NiCO+3H2syngasC{{H}_{4}}+{{H}_{2}}O\xrightarrow{1273K,Ni}\underset{syngas}{\mathop{CO+3{{H}_{2}}}}\,
(ii)- Catalytic oxidation: When a mixture of methane and oxygen (9:1 by volume) at a pressure of 100 atmospheres is passed through a copper tube at 573 K, methanol is formed.
2CH4+O2Cu tube100atm,573K2CH2OH2C{{H}_{4}}+{{O}_{2}}\xrightarrow[Cu\text{ }tube]{100atm,573K}2C{{H}_{2}}OH.

Note: The process of complete oxidation of alkane is accompanied by the liberation of a large amount of heat, therefore, alkanes which are the constituents of LPG, gasoline, kerosene oil, and diesel are used as fuels.