${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\left(g\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)?$

Magnitude of volume change if 30 ml of ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{g}\right)$ is burnt with excess amount of oxygen, is :

 The equation displayed is

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{OH}}_{\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}⟶{\mathrm{CO}}_{2\left(\mathrm{g}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$
Following a balanced equation

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{OH}}_{\left(\mathrm{g}\right)}+7{\mathrm{O}}_{2\left(\mathrm{g}\right)}⟶6{\mathrm{CO}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

∴ Change in volume =Product volume − reactant volume

⇒(180+90)−(210+30)

 ⇒270−240

 ⇒30 mL