Calculate the volume of HCl gas formed and chlorine gas required when 40 ml of methane reacts completely with chlorine at STP.

CH₄ + 2Cl₂ ⟶ CH₂Cl₂ + 2HCl

$\begin{matrix} \text{CH}$4 & + & 2\text{Cl}2 & \longrightarrow & \text{CH}2\text{Cl}2 & + & 2\text{HCl} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1 \text{ vol.} & : & 2 \text{ vol.}\\ \end{matrix}

volume of HCl gas formed = ?

[By Gay Lussac's law]

1 Vol of methane produces = 2 Vol. HCl

∴ 40 ml of methane produces = 80 ml HCl

volume of chlorine gas required = ?

For 1 Vol of methane = 2V of Cl₂ required

∴ for 40 ml of methane = 40 x 2 = 80 ml of Cl₂ is required.

Hence, volume of HCl gas formed = 80 ml and chlorine gas required = 80 ml