2Al(s)+3Cl2 = 2AlCl3

Question

Limiting Reactant is Cl with 0.329mol.

What is the maximum mass of aluminum chloride that can be formed when reacting 30.0g of aluminum with 35.0g of chlorine?

Answer 1

35g of Al*(1mole of Cl/35.45g)= mole of Cl

3 moles of Cl= 2 moles of AlCl3

moles of Cl *(2 moles of AlCl3/3 moles of Cl)= moles of AlCl3

moles of ALCl3 *(133.34 g of AlCl3/mol)= mass of AlCl3