Question

How many moles of oxygen are required to react completely with 16.5 mol Al?

Answer 1

12.4 mol O2

Step-by-step explanation:

4Al + 3O2 -----> 2Al2O3

from reaction 4 mol 3 mol

given 16.5 mol x mol

x = 16.5*3/4 =12.4 mol O2