first, balance the equation:
2Al + 3Cl2 → 2AlCl3
Now, how many moles of each reagent do you have?
114g Al = 4.225 moles
186g Cl2 = 2.657 moles
so, if all the Al is used up, you will need 4.225(3/2) = 6.338 moles of Cl2
You don't have that, so clearly the Cl2 limits the reaction
In a reaction, 114 grams of Al reacts with 186 grams of Cl2. What is the limiting reactant?
Al + Cl2 ----> AlCl3
1 answer
1 answer