The balanced equation tells us we need 2 moles of Al for every 3 moles of Cl2. So let's see if we have enough, or more than enough, Cl2 given that we have 12.6 g of Al. Determine the moles of each reactant that we have available:
Moles is obtained by dividing the mass by the molar mass [g/(g/mole) = moles]. The required molar ratio of chlorine to aluminum is (1.5 moles Cl2)/(mole Al), based on the balanced equation. Multiply the 1.5 times the moles of Al (0.474 moles Al) to determine how many moles of Cl2 would need to be present to fully react with the Al.
This tells us we would need 0.7111 moles of chlorine to fully react, but we only have 0.551 moles of Cl2. We'll run out of chlorine once the Al is consumed, so chlorine is the limiting reagent. In fact, we can determine how much aluminum remains after the chlorine is consumed. If all of the chlorine reacts, it would consume:
0.551 moles Cl2 available x (2 moles Al/3 moles Cl2) = 0.316 moles Al.
Remaining moles of Al = (0.474 - 0.316) = 0.158 moles Al remaining.