Question

When $10\mathrm{mL}$ of ${\mathrm{H}}_2$​ and $12.5\mathrm{mL}$ of ${\mathrm{Cl}}_2$ are allowed to react, the final mixture contains under the same conditions.

Answer 1

The final mixture under some conditions :

1. Reaction is taking place at STP condition (assumption) so that the $10\mathrm{mL}$ of ${\mathrm{H}}_2$ and $12.5\mathrm{mL}$ of ${\mathrm{Cl}}_2$is just equal to $0.45$ and $0.55$ millimoles of ${\mathrm{H}}_2$ and ${\mathrm{Cl}}_2$ respectively.
2. $2\mathrm{HCl}$ is basically the product that is obtained when ${\mathrm{H}}_2$ and ${\mathrm{Cl}}_2$ reacts.
3. So, it can be said that $1\mathrm{mole}$ of ${\mathrm{H}}_2$ reacts with $1\mathrm{mole}$ of ${\mathrm{Cl}}_2$ further to form the $2\mathrm{moles}$ of $\mathrm{HCl}$.
4. Similarly, when $0.45\mathrm{millimoles}$ of ${\mathrm{H}}_2$ reacts with $0.45\mathrm{millimoles}$ of ${\mathrm{Cl}}_2$ and so $0.9\mathrm{millimoles}$ of $\mathrm{HCl}$ is formed and the unreacted ${\mathrm{Cl}}_2$ is $0.1\mathrm{millimoles}$.
5. Therefore, the reaction mixture will contain $0.10\mathrm{millimoles}$ of ${\mathrm{Cl}}_2$ and $0.9\mathrm{millimoles}$ of $\mathrm{HCl}$
6. Hence, At STP, this is equal to $20.16\mathrm{mL}\mathrm{of}\mathrm{HCl}$ and $2.24\mathrm{mL}\mathrm{of}{\mathrm{Cl}}_2$.