You are given 33.0 g of aluminum and 38.0 g of chlorine gas
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well since aluminum is in excess it doesn't matter so we look at chlorine gas which will determine how much aluminum chloride can be made since its not in excess. First we have to do a balance eq. which is to balance the charges: Al is +3 while Cl -1 so we end up with AlCl3 then balance the amount of elements by placing #'s in the front of the elements 2AlCl3 3Cl2 + 2Al -- 2AlCl3 so... since Cl2 is not in excess we calcutlate how much AlCl3 can be produced 1) use (mol/g) from periodic table remember times by 2 since 2 Cl then once in mols you can do pretty much anything! 2)so now that your in mols use mol to mol ratio to convert to what you want with the help of the balanced eq. 38.0g Cl2 (1mol Cl2/ 2(35. something grams))(2mol AlCl3/ 3mol Cl2) and tah dah you have mols of AlCl3! and it will have 3 sig figs!
