Chemistry - Eudiometry Question with Solution | TestHub

$\mathrm{CH}+$$2{\mathrm{O}}_2⟶{\mathrm{CO}}_2+{\mathrm{H}}_2{\mathrm{O}}^0$

at $\mathrm{t}=0$ x ml $(80-\mathrm{x})\mathrm{ml}$ 0 0

at $\mathrm{t}=\mathrm{t}$ 0 $(80-\mathrm{x}-2\mathrm{x})\mathrm{ml}$ x ml 0 ←When CH4 is limiting

Now, when CH₄ is limiting:

$\text{ Contraction }=80-[80-3x+x]$

$40=80-80+2x$

$x=20\Rightarrow {\mathrm{CH}}_4\to 20\mathrm{ml}$

${\mathrm{O}}_2\to 60\mathrm{ml}$

When ${\mathrm{O}}_2$ is limiting:

$40=80-\left[x-\left(\frac{80-x}{2}\right)+\frac{80-x}{2}\right]$

$x=40\Rightarrow {\mathrm{CH}}_4=40\mathrm{ml};{\mathrm{O}}_2=40\mathrm{ml}$