a) m Al = 5,4g
n Al = 5,4/27 = 0,2 moli
m Cl2 = 30g
n Cl2 = 30/71 = 0,422 moli
2Al + 3Cl2 => 2AlCl3
2 moli Al..................3 moli Cl2
0,2 moli Al...............x = 0,3 moli Cl2
2 moli Al..................3 moli Cl2
x moli Al.................0,422 moli Cl2
x = 0,281 moli Al
Din cele doua ecuatii de 3 simple se observa clorul ca fiind in exces.
n moli Cl2 exces = 0,422-0,3 = 0,122 moli
m Cl2 exces = 8,7g
b) 2 moli Al................2×133,5g AlCl3
0,2 moli Al.......................x g AlCl3
x = 2×13,35g = 26,7g AlCl3
