Question

What is the greatest amount of AlCl3 (in grams) that can be made with 114 grams of Al and 186 grams of Cl2? Which is the limiting reactant? Which reactant is in excess? (Al=26.98) (Cl=35.46) Al + Cl2 --> AlCl3

Answer 1

The balanced equation is

2Al + 3Cl₂ --> 2AlCl₃

$n(Al)=\frac{114\ g}{26.98\ g/mol}=4.225\ mol$

$n(Cl_2)=\frac{186\ g}{70.92\ g/mol}=2.623\ mol$

According to the equation, 2 moles of Al react with 3 moles of Cl₂. Actually, there are less moles Cl₂ than Al, which obviously means that Cl₂ is the limiting reactant.

$m(AlCl_3)=2.623\ mol(Cl_2)\times\frac{2\ mol(AlCl_3)}{3\ mol(Cl_2)}\times\frac{133.34\ g(AlCl_3)}{1\ mol(AlCl_3)}=233\ g$

Answer: Cl₂ - limiting; Al - excess; 233 g AlCl₃