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Chemistry

If 6 L of hydrogen and 4 L of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.

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Answer

[By Lussac's law]

H2+Cl2⟶2HCl1 vol.:1 vol.⟶2 vol.\begin{matrix} \text{H}2 & + & \text{Cl}2 &\longrightarrow & 2\text{HCl} \ 1 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}

To calculate the amount of hydrogen used:

Cl2:H21 vol.:1 vol.4 lit.:x\begin{matrix}\text{Cl}2 & : & \text{H}2 \ 1 \text{ vol.} & : & 1 \text{ vol.} \ 4 \text{ lit.} & : & \text{x} \end{matrix}

∴x=11×4=4 lit.\therefore x = \dfrac{1}{1} \times 4 = 4 \text{ lit.}

Remaining hydrogen = 6 - 4 = 2 lit.

To calculate the amount of HCl :

Cl2:HCl1 vol.:2 vol.4 lit.:x\begin{matrix}\text{Cl}_2 & : & \text{HCl} \ 1 \text{ vol.} & : & 2 \text{ vol.} \ 4 \text{ lit.} & : & \text{x} \end{matrix}

∴x=21×4=8 lit.\therefore x = \dfrac{2}{1} \times 4 = 8 \text{ lit.}

Hence, after reaction 8 lit of HCl is formed which dissolves in water and 2 lit of hydrogen is left.

∴ 2 lits. of hydrogen is left.

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