Question

A sample of gaseous hydrocarbon occupying 1.12 litre at NTP, when completely burnt in air produced 2.2 g $C{O}_2$ and 1.8 g $H_{2}O$. Calculate the mass of hydrocarbon taken and the volume of $O_2$ at NTP required for its combustion.

• A. 1.14 L
• B. 2.24 L
• C. 2.4 L
• D. 2.8 L

Answer 1

The correct option is B

2.24 L

You know that a hydrocarbon is made up of C and H

$\therefore$ $C_a{H}_b$ + (a + $\frac{b}{4})O_2$ $\to$ a $C{O}_2$ + $\frac{b}{2}$ $H_{2}O$ (I) is the combustion of the hydrocarbon.

1 mole of $C_a{H}_b$ = (44a) g of $C{O}_2$

1.12 L of $C_a{H}_b$ = ($\frac{1.12\times 44a}{22.4}$) g of $C{O}_2$

From question $\frac{1.12\times 44a}{22.4}$ = 2.2g $\Rightarrow$ a = 1

Again, 22.4 L of $C_a{H}_b$ = ($\frac{b}{2}$ $\times$ 18)g of $H_{2}O$

1.12 of $C_a{H}_b$ = $\frac{1.12\times \frac{b}{2}\times 18}{22.4}$

From question, $\frac{1.12\times \frac{b}{2}\times 18}{22.4}$ = 1.8 $\Rightarrow$ b = 4

$\therefore$ The hydrocarbon of methane ($C{H}_4$)

$\therefore$ Number of moles of oxygen = ( 1 + $\frac{4}{4}$) = 2 moles.

From reaction,

22.4 L of $C{H}_4$ = 2 $\times$ 22.4 L of $O_2$

1.12 L of $C{H}_4$ = $\frac{1.12\times 2\times 22.4}{22.4}$

1.12 L of $C{H}_4$ = 2.24L of $O_2$