Question

Q8. Calculate the volume of $\mathrm{HCl}$ gas formed and chlorine gas required when $40ml$ of methane reacts completely with chlorine at STP.

$\underset{\mathrm{methane}}{{\mathrm{CH}}_4}+\underset{\mathrm{chlorine}}{2{\mathrm{Cl}}_2}\to \underset{\mathrm{dichloro}\mathrm{methane}}{{\mathrm{CH}}_2{\mathrm{Cl}}_2}+\underset{\mathrm{hydrochloric}\mathrm{acid}}{2\mathrm{HCl}}$

Answer 1

• The given equation:

$\underset{1vol}{{\mathrm{CH}}_4\left(\mathrm{g}\right)}+\underset{2vol}{2{\mathrm{Cl}}_2\left(\mathrm{g}\right)}\to \underset{1vol}{{\mathrm{CH}}_2{\mathrm{Cl}}_2\left(\mathrm{g}\right)}+\underset{2vol}{2\mathrm{HCl}\left(\mathrm{g}\right)}$

• calculating the volume of $\mathrm{HCl}$

$1vol$of ${\mathrm{CH}}_4$ produces $\mathrm{HCl}$$=2$ vol(at STP)

Therefore, $40$vol of ${\mathrm{CH}}_4$ produces $\mathrm{HCl}$ at STP$$\begin{gathered} =2\times 40 \\ =80ml \end{gathered}$$

• The resulting volume of $\mathrm{HCl}$ produced at STP$80ml$

(Vol of ${\mathrm{Cl}}_2$required$$\begin{gathered} =40\times 2 \\ =80ml \end{gathered}$$)