40 cm³ of methane reacts with chlorine. The reaction for the same is

CH₄ + 2Cl₂ ⟶ CH₂Cl₂ + 2HCI

Find

(a) The volume of chlorine gas used.

(b) The volume of hydrogen chloride gas formed.

(c) How is hydrogen chloride gas dried?

$\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}

(a) Volume of chlorine gas used = ?

For 1 Vol of methane = 2V of Cl₂ required

∴ For 40 cm³ of methane = 40 x 2 = 80 cm³ of Cl₂ is required.

(b) Volume of HCl gas formed = ?

[By Gay Lussac's law]

1 Vol of methane produces = 2 Vol. HCl

∴ 40 cm³ of methane produces = 80 cm³ HCl

Hence, volume of HCl gas formed = 80 cm³ and chlorine gas required = 80 cm³

(c) Hydrogen chloride gas is dried by passing through a washer bottle containing concentrated sulphuric acid.