Răspuns:
Explicație:
MAl(OH)3= AAl + 3.16 + 3AH
=27 + 48 +3
=78--------> 78g/moli
MAl2(SO4)3= 2.27 + 3.32 + 12.16
=54 + 96 +192
=342 ------> 342g/moli
5.
H2SO4
M= 2+32 +4.16=98------> 98g/moli
98g H2SO4----------2gH--------32g S-------64g O
100g------------------------x-----------y------------------z
x=100 . 2 : 98=2,04%H
y=100 . 32 : 98=32,65% S
z=100 .64 : 98=65,31%O
Fe2O3
M=2.56+ 3.16=160-------> 160g/moli
160g Fe2O3----------112g Fe---------48gO
100g----------------------X----------------------Y
X= 100.112 : 160=70% Fe
Y= 100 .48 : 160=30% O