The rate of formation of oxygen in the reaction 2N₂O₅(g)$\rightarrow$ 4NO₂(g)+ O₂(g)Is 2.28 (mol O₂).L^-¹.s^-¹.What is the rate of formation of NO₂? A) 0.57 (mol NO₂).L^-¹.s^-¹ B) 9.12 (mol NO₂).L^-¹.s^-¹ C) 2.28 (mol NO₂).L^-¹.s^-¹ D) 1.14 (mol NO₂).L^-¹.s^-¹ E) 4.56 (mol NO₂).L^-¹.s^-¹

Question

Quizplus · Accepted Answer

The rate of formation of $NO$ is four times the rate of formation of $O_2$ because for every molecule of $O_2$ formed, four molecules of $NO$ are produced according to the stoichiometry of the reaction. Therefore, $2.28 \times 4 = 9.12$ mol $NO$.

The Rate of Formation of Oxygen in the Reaction 2N2O5(g) →\rightarrow→

The Rate of Formation of Oxygen in the Reaction
2N₂O₅(g) $\rightarrow$