If you had excess aluminum, how many moles of aluminum chloride could be produced from 17.0 g of chlorine gas, cl2?

  The moles  of aluminum  chloride could be produced  from 17.0g chloride  gas  is  0.159  moles

       calculation

write  the equation for formation  of  aluminum chloride
that is  2Al + 3Cl2→ 2 AlCl3

find  the  moles of Cl2 used  

moles= mass  of Cl2 /molar  mass of Cl2 (  35.5 x2= 71 g/mol)
17g/71 g  = 0.239  moles

by  use of mole ratio  between  Cl2 to AlCl3  which is 3:2  the moles of   AlCl3 is therefore=  0.239 x2/3 = 0.159  moles