What is the limiting reactant for the reaction below given that you start with 0.37 mol of Al and 0.59 mol of O2? Reaction: 4Al + 3O2 → 2Al2O3
Responses
To determine the limiting reactant, we have to compare the amount of moles of aluminum and moles of oxygen to see which one will run out first.
From the balanced equation, we see that it takes 4 moles of aluminum for every 3 moles of oxygen to produce 2 moles of aluminum oxide.
Calculate the moles of Al2O3 that can be produced from 0.37 mol of Al:
0.37 mol Al * (2 mol Al2O3 / 4 mol Al) = 0.185 mol Al2O3
Calculate the moles of Al2O3 that can be produced from 0.59 mol of O2:
0.59 mol O2 * (2 mol Al2O3 / 3 mol O2) = 0.393 mol Al2O3
Since we can only produce 0.185 mol of Al2O3 from 0.37 mol of Al and 0.393 mol of Al2O3 from 0.59 mol of O2, the limiting reactant is aluminum (Al).