Please help, I'm desperate!?

Well, you won't use all of the chlorine in the reaction, because the ratio of Al to Cl2 is only 2 to 3, which is a lot more than 5 to 34. The reaction will only use as much chlorine as will react with the aluminum. With a ratio of 2Al to 3Cl2, 5 moles of Al will need 7.5 moles of Cl2. Figure out how much each of those weighs, using the molecular weights of the two elements. Don't forget to count twice for the two atoms of chlorine. That's how much AlCl3 you'll get.