A 2L sample of a gaseous hydrocarbon is burnt in excess oxygen. The only products of the reaction are 8 L of CO 2 ( g ) and 10 L of H 2 O ( g ) , all at 10 0 ∘ C and 1 atm pressure. The formula of

Question

A 2L sample of a gaseous hydrocarbon is burnt in excess oxygen. The only products of the reaction are 8 L of CO 2 ​ ( g ) and 10 L of H 2 ​ O ( g ) , all at 10 0 ∘ C and 1 atm pressure. The formula of the hydrocarbon is -

TestHub · Accepted Answer

The combustion of a hydrocarbon follows the general equation:

$C_{x} H_{y} (g) + (x + y /4) O_{2} (g) \to x C O_{2} (g) + (y /2) H_{2} O (g)$

According to Avogadro's Law, for gases at the same temperature and pressure, the ratio of volumes is equal to the ratio of moles.

Given:

Volume of hydrocarbon ( $C_{x} H_{y}$ ) = 2 L

Volume of $C O_{2}$ = 8 L

Volume of $H_{2} O$ = 10 L

From the stoichiometry of the reaction:

Moles of $C_{x} H_{y}$ : Moles of $C O_{2}$ = 1 : x

Volume of $C_{x} H_{y}$ : Volume of $C O_{2}$ = 1 : x

$2 L : 8 L = 1 : x \Rightarrow x = 4$

Moles of $C_{x} H_{y}$ : Moles of $H_{2} O$ = 1 : y/2

Volume of $C_{x} H_{y}$ : Volume of $H_{2} O$ = 1 : y/2

$2 L : 10 L = 1 : y /2 \Rightarrow y /2 = 5 \Rightarrow y = 10$

Thus, the formula of the hydrocarbon is $C_{4} H_{10}$ .

The final answer is $C_{4} H_{10}$ .

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