Răspuns:
Explicație:
1.
Al2 (SO4)3
CaO
NaCl
MgCO3
2.
MH2CO3 =2 + 12 + 3.16=62 -----> 62g/moli
MAl2O3= 2.27 + 3.16=102-------> 102g/moli
MFe(OH)2=56 +2.16 + 2=90-----> 90g/moli
MMg3(PO4)2= 3.24 + 2.31 + 8.16=262-------> 262g/moli
3.
MFe(OH)2 = 56 + 2.16 +2=90 ---> 90g/moli
n= m : M
m= n . M
m= 2moli . 90g/moli=180g Fe(OH)2
4.
n= m : M
n= 270g : 90g/moli= 3moli Fe(OH)2
5.
168g Fe
in ce cantit. de Fe(OH)2
M Fe(OH)2 = 56 + 2.16 + 2= 90------> 90g/moli
90g Fe(OH)2 --------56g Fe
xg Fe(OH)2-----------------168gFe
x= 90 . 168 : 56 = 270 g Fe(OH)2
6.
MFe(OH)2=56 + 2.16 + 2= 90------> 90g/moli
90 g Fe(OH)2-------32g O
270g Fe(OH)2----------x g O
x= 270 . 32 : 90 = 96 g O
MH2CO3= 2 +12 + 3.16=62------> 62g/moli
62g H2CO3---------48 g O
yg H2CO3----------96 g O
y= 62 . 96 : 48 = 124 g H2CO3
7.
1mol Fe(OH)2----------6,023.10²³molecule Fe(OH)2
2moli----------------------------x
x=2 . 6,023.10²³molecule= 12,046 . 10²³
