Question

va rog frumos sa ma ajutati, multumesc!
dau coroana​

Answer 1

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