Question

15cm3 of a gaseous hydrocarbon was mixed with 120cm3 of oxygen which was in excess. The mixture was exploded, and after cooling, the residual volume was 105cm3. On adding conc. KOH, the volume reduced to 75cm3. What's the molecular formular of the HC?

Answer 1

The final volume of the reaction between $CO_2$ and $O_2$ was $105 cm^3$

$105cm^3-50cm^3=50cm^3$

$50cm^3$ Is the remaining volume after the reaction.

The volume allowed for the reaction is;

$120cm^3-50cm^3=70cm^3$

The coefficients for the reaction are;

$2C_x+H_y+7O_2\to4CO_2+H_2O$

These are coefficients for the ratios of the gas to volume ratio. Which are simplified to;

$20(C_xH_y):70(CO_2);55(CO_2)$

The equation for oxygen is;

$2C_xH_y+7O_2\to4CO_2+6H_2O$

$\therefore$ $x=2, y=8$

The compound has a molecular formula of $C_2H_8$

The compound is methane.