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