How many moles of oxygen are produced when 26.5 mol of aluminum oxide are decomposed?

The balanced chemical reaction is:

2Al2O3 = 4Al +3 O2

We are given the amount of the reactant. This will be the starting point of our calculations.

26.5 mol Al2O3 ( 3 mol O2 / 2 mol Al2O3 ) = 39.75 mol O2 needed

Thus, 0.014 mol of I2 is needed to form the given amount of NI3.

the balanced chemical equation for the decomposition of Al₂O₃ is as follows

2Al₂O₃ --> 4Al + 3O₂

stoichiometry of Al₂O₃ to O₂ is 2:3

when 2 mol of Al₂O₃ decomposes - 3 mol of O₂ are formed

therefore when 26.5 mol of Al₂O₃ decomposes - 3/2 x 26.5 mol = 39.75 mol

the number of moles of O₂ formed are 39.75 mol

im not exactly sure but i think its b!