Help Molar / Mass Problem

Question

A mass of 45 grams of Aluminum are used up by the reaction, how many moles of Aluminum Chloride will be produced

Answer 1

Assuming the reaction is between solid aluminum and gaseous chlorine, the balance equation would be...

2Al(s) + 3Cl₂(g) ===> 2AlCl₃(s)

moles of Al used = 45 g Al x 1 mole Al/27 g = 1.67 moles Al = 1.7 moles Al

Assuming that Cl₂(g) is not limiting in the reaction, the formation of AlCl₃ will depend only on Al present.

Thus, moles AlCl₃ produced = 1.7 moles Al x 2 moles AlCl₃/2 moles Al = 1.7 moles AlCl₃ produced