Question

Two moles of a hydrocarbon for complete combustion consume 13 moles of oxygen. When 1 mole of it is subjected to complete combustion, the number of moles of water vapour formed is one mole more than the number of moles of ${\mathrm{CO}}_2$ formed. Identify the formula of hydrocarbon.

• A. ${\mathrm{C}}_4{\mathrm{H}}_{10}$
• B. ${\mathrm{C}}_5{\mathrm{H}}_{12}$
• C. ${\mathrm{C}}_5{\mathrm{H}}_{10}$
• D. ${\mathrm{C}}_4{\mathrm{H}}_8$

Answer 1

The correct option is A

${\mathrm{C}}_4{\mathrm{H}}_{10}$

Explanation for the correct option:

(A)

• Hydrocarbons on combustion produce Carbon dioxide (${\mathrm{CO}}_2$) and Water (${\mathrm{H}}_2\mathrm{O}$).
• The general equation for the combustion of Alkanes is given below:
• ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}+\frac{(3\mathrm{n}+1)}{2}{\mathrm{O}}_2\to \mathrm{n}{\mathrm{CO}}_2+(\mathrm{n}+1){\mathrm{H}}_2\mathrm{O}$
• The general equation for the combustion of Alkenes is given below:
• ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}}+\frac{\left(3\mathrm{n}\right)}{2}{\mathrm{O}}_2\to \mathrm{n}{\mathrm{CO}}_2+\mathrm{n}{\mathrm{H}}_2\mathrm{O}$
• The condition given in the question 'when 1 mole of it is subjected to complete combustion, the number of moles of water vapor formed is one mole more than the number of moles of ${\mathrm{CO}}_2$ formed' clearly shows that the hydrocarbon is Alkane.

Now, for Butane (${\mathrm{C}}_4{\mathrm{H}}_{10}$), putting the values of n in the general equation, we get:

$\underset{\mathrm{Butane}}{{\mathrm{C}}_4{\mathrm{H}}_{10}}\left(g\right)+\underset{\mathrm{Oxygen}}{\frac{13}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{4{\mathrm{CO}}_2\left(g\right)}+\underset{\mathrm{Water}}{5{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

When we take the 2 moles of Butane is taken then 13 moles of Oxygen is consumed to produce 8 moles of Carbon dioxide and 10 moles of Water will be produced. The reaction is given below:

$\underset{\mathrm{Butane}}{2{\mathrm{C}}_4{\mathrm{H}}_{10}}\left(g\right)+\underset{\mathrm{Oxygen}}{13{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{8{\mathrm{CO}}_2\left(g\right)}+\underset{\mathrm{Water}}{10{\mathrm{H}}_2\mathrm{O}}\left(l\right)$

Explanation for the incorrect option:

(B) For Pentane (${\mathrm{C}}_5{\mathrm{H}}_{12}$), putting the value of n in the general reaction:

$\underset{\mathrm{Pentane}}{{\mathrm{C}}_5{\mathrm{H}}_{12}\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{8{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{5{\mathrm{CO}}_2\left(g\right)}+\underset{\mathrm{Water}}{6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

When we take the 2 moles of pentane is taken then 16 moles of Oxygen is consumed to produce 10 moles of Carbon dioxide and 12 moles of Water will be produced. The reaction is given below:

$\underset{\mathrm{Pentane}}{2{\mathrm{C}}_5{\mathrm{H}}_{12}\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{16{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{10{\mathrm{CO}}_2\left(g\right)}+\underset{\mathrm{Water}}{12{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

As Pentane does not follow the condition given in the question.

So, Option (C) and Option (D) are incorrect because they are alkenes.

Therefore, the correct option is (A).