The ratio of active masses of , , and present in a single gaseous mixture inside a closed vessel is: — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Moles of $CO_2 = \frac{22}{44} = 0.5$. Moles of $H_2 = \frac{3}{2} = 1.5$. Moles of $N_2 = \frac{7}{28} = 0.25$. Since they are in the same volume $V$, the ratio of their active masses is the ratio of their moles: $0.5 : 1.5 : 0.25$. Dividing by $0.5$, we get $1 : 3 : 0.5$.