Question

determine limiting reagent if 34g of aluminium reacts with 39g of chloride also find the amount of alluminium chloride

Answer 1

Answer:

2 Al + 3 Cl2 ==> 2 AlCl3 Balanced Equation

moles Al present = 16.5 g Al x 1 mole Al/26.98 g = 0.612 moles Al

moles Cl2 prsent = 39.2 g Cl2 x 1 mole Cl2/35.45 g = 1.11 moles Cl2

Dividing each reactant by it's coefficient in the balanced equation one obtains:

0.612 moles Al/2 = 0.306 mols Al

1.11 moles Cl2/3 = 0.37 moles Cl2

LIMITING REACTANT IS Al

Theoretical Yield will thus depend on moles of Al and the mole ratio of AlCl3 : Al which is 1 : 1 (see balanced equation)

0.306 moles Al x 2 moles AlCl3/2 moles Al = 0.306 moles AlCl3 = theoretical yield (in moles)

0.306 moles AlCl3 x 133.3 g/mole = 40.8 g AlCl3 = theoretical yield (in grams)