Respuesta :

alright! well let’s we specify a molar quantity of
0.44

m
o
l
with respect to aluminum sulfate….i.e. a mass of
150.5

g
...and in this quantity there are TWO EQUIV aluminum….i.e. a mass of
2
×
0.44

m
o
l
×
27.0

g
=
23.8

g
...

Answer:

0.88  ⋅mol   of aluminum metal in this molar quantity of salt....

Explanation: