Chemistry - Eudiometry Question with Solution | TestHub

$\mathrm{CH}+2{\mathrm{O}}_2⟶{CO}_2+$

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

$at\mathrm{t}=\mathrm{t}0(80-\mathrm{x}-2\mathrm{x})\mathrm{ml}$ $\mathrm{xml}0$ $\leftarrow -When{\mathrm{CH}}_{4}islimiting.$

Now, When CH₄ is limiting

$\therefore Contraction=80-[80-3x+x]$

$40=80-80+2x$

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

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

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

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

$x=40-80\Rightarrow 40\mathrm{ml}$

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