Chemistry - Mole Concept Question with Solution | TestHub

${\mathrm{CH}}_4+ 2{\mathrm{O}}_2⟶ {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$

$\frac{100}{16} \frac{208}{32}$

$$\begin{gathered} \frac{\mathrm{Number} \mathrm{of} \mathrm{moles}}{\mathrm{stoichiometric} \mathrm{coefficient}} \mathrm{for} {\mathrm{CH}}_4=\frac{6.25}{1} \\ \frac{\mathrm{Number} \mathrm{of} \mathrm{moles}}{\mathrm{stoichiometric} \mathrm{coefficient}} \mathrm{for} {\mathrm{O}}_2=\frac{6.5}{2}=3.25 \end{gathered}$$

The reagent with lower ratio of number of moles to stoichiometric coefficient is called limiting reagent.

Hence, ${\mathrm{O}}_2$ is limiting reagent

Limiting reagent decides the amount of product,

As per the equation 2 moles of ${\mathrm{O}}_2$ gives rise to 1 mole of ${\mathrm{CO}}_2$, So 6.5 moles of ${\mathrm{O}}_2$ gives $\frac{6.5}{2}$ moles of ${\mathrm{CO}}_2$

$\Rightarrow {\mathrm{nCO}}_2=\frac{6.5}{2}$

So mass of ${\mathrm{CO}}_2=\frac{6.5}{2}\times 44=143\mathrm{gm}$