Answer:
Part A:
Molar mass of Al = 26.98 g/mol
mass of Al = 33.0 g
we have below equation to be used:
number of mol of Al,
n = mass of Al/molar mass of Al
=(33.0 g)/(26.98 g/mol)
= 1.223 mol
Part B
Molar mass of Cl2 = 70.9 g/mol
mass of Cl2 = 38.0 g
we have below equation to be used:
number of mol of Cl2,
n = mass of Cl2/molar mass of Cl2
=(38.0 g)/(70.9 g/mol)
= 0.536 mol
Part B
we have the Balanced chemical equation as:
2 Al + 3 Cl2 ---> 2 AlCl3
2 mol of Al reacts with 3 mol of Cl2
for 1.2231 mol of Al, 1.8347 mol of Cl2 is required
But we have 0.536 mol of Cl2
so, Cl2 is limiting reagent
we will use Cl2 in further calculation
Molar mass of AlCl3 = 1*MM(Al) + 3*MM(Cl)
= 1*26.98 + 3*35.45
= 133.33 g/mol
From balanced chemical reaction, we see that
when 3 mol of Cl2 reacts, 2 mol of AlCl3 is formed
mol of AlCl3 formed = (2/3)* moles of Cl2
= (2/3)*0.536
= 0.3573 mol
