112 cm³ of H₂S (g) is mixed with 120 cm³ of Cl₂ (g) at STP to produce HCl (g) and sulphur (s). Write a balanced equation for this reaction and calculate

(i) the volume of gaseous product formed

(ii) composition of the resulting mixture

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

(i) At STP,

1 mole gas occupies = 22.4 L.

1 mole H₂S gas produces = 2 moles HCl gas,

∴ 22.4 L H₂S gas produces

= 22.4 × 2

= 44.8 L HCl gas.

Hence, 112 cm³ H₂S gas will produce

= 112 × 2

= 224 cm³ HCl gas.

Hence, 224 cm³ HCl gas is produced.

(ii) 1 mole H₂S gas consumes = 1 mole Cl₂ gas.

Hence, 22.4 L H₂S gas consumes = 22.4 L Cl₂ gas at STP.

∴ 112 cm³ H₂S gas consumes = 112 cm³ Cl₂ gas.

120 cm³ - 112 cm³ = 8 cm³ Cl₂ gas remains unreacted.

Hence, the composition of the resulting mixture is 224 cm³ HCl gas + 8 cm³ Cl₂ gas.