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Sagot :
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
...
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:
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