Aluminum reacts with oxygen to form aluminum oxide. (a) How many moles of O2 are needed to react with 2.09 mol of aluminum?
1.57 mol O2 4Al + 3O2-- 2Al2O3 2.09 * (3/4) = 1.5675 = 1.57 mol O2 (with sig figs)
from the equation you have mols of Al mandatory=2 mols of FeO mandatory=3 as a result a million mole of FeO woulde require 3/2 mols of Al once you have a million.2 mols of FeO you'll want a million.2*3/2= a million.8mols i'm hoping you recognize
The first step is to write a balanced equation. Aluminum oxide is Al2O3 because aluminum commonly takes on a charge of 3+, while oxygen tends to form ions with 2- charge. 4 Al + 3 O2 -- 2 Al2O3 Now, all you have to do is use the mole ratio (which comes from the balanced equation) to solve this problem. You are given moles of aluminum and asked to find moles of oxygen, so the mole ratio you want is that which relates those two, which would be 3/4 moles O2/moles Al (because those are the coefficients). So you just have to multiply the moles of Al by this ratio to find the number of moles of O2: (2.09 mol Al) * (3/4 mol O2/mol Al) = 1.57 mol O2
