Question

10 mlof gaseous hydrocarbon on combustion give 40 ml of co2 and 50 ml of h2o.The hydrocarbon is

Answer 1

Explanation:

Provided they are all gases then the formula is going to be C4H10.

Answer 2

${ \bold{ \: Let \: the \: hydrocarbon \: be \: C_xH_y.}}$

It is a combustion reaction which means it is reacting with Oxygen.

${ \bold{ C_xH_y + O_2 \: –› \: CO_2 + H_20}}$

Balance the equation,

${ \bold{ \:C_xH_y + { \green{ \: (2x + \frac{y}{2}) }}O_2 \: –› \: { \green{X}}CO_2 + { \green{ \frac{y}{2}}} H_2O }} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (10 \: ml)\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (40 \: ml) \: \: \: \ \: \: (50 \: ml)\: \: \: \: \: \: \: \: \:$

We know that,

According to volume-volume analysis,

$\: \longrightarrow \: { \bold{ \frac{Volume_{C_xH_y}}{1} = \frac{Volume_{CO_2} }{X} }} \\ \\ \: \longrightarrow \: { \bold{ \frac{10}{1} = \frac{40}{X} }} \\ \\ \: \longrightarrow \: { \boxed{ \bold{ X = 4}}}$

$\: \longrightarrow \: { \bold{ \frac{Volume_{C_xH_y}}{1} = \frac{Volume_{H_2O} }{ \frac{y}{2} } }} \\ \\ \: \longrightarrow \: { \bold{ \frac{10}{1} = \frac{50}{ \frac{y}{2} } }} \\ \\ \: \longrightarrow \: { \bold{ 10 \times \frac{y}{2} = 50}} \\ \\ \: \longrightarrow \: { \boxed{ \bold{ y = 10 }}}$

Hence,

$\: { \green{ \boxed{ \tt{Molecular \: \: Formula = C_4H_{10}}}}}$