# Tag Info

The excess of $\ce{O2}$ is ${7.32~ mol}$. So the excess air is : ${7.32 ~mol· (1 + 3.76) = 34.84~ mol}$. The total amount of air is : ${a·4.76 = 19.53~ mol ·4.76 = 92.96~ mol}$. The percentage of excess air is : $34.84/92.96 = 0.3748 = 37.48$%