Răspuns:
Explicație:
1.
60 g Mg , 10% imp.
react. cu O2
g, moli produs de react.
-se afla puritatea Mg
p=100 - 10= 90%
-se afla masa pura de Mg
mp= p . mi : 100
mp= 90 . 60 : 100= 54g Mg pur
54g xg
2Mg + O2= 2 MgO
2.24g 2. 40g
x= 54 . 80 : 48=90 g MgO
n= 90g : 40g/moli= 2,25 moli MgO
MMgO= 24 + 16=40------> 40g/moli
2.
50 g Ca , p= 80%
react. cu o sol. H2SO4, c= 20%
ms H2SO4
masa de sare
Vgaz
-se afla mp de Ca
mp= p . mi : 100
mp= 80 .50 : 100=40g Ca pur
40g xg yg zg
Ca + H2SO4 = CaSO4 + H2
40g 98g 136g 2g
x= 40 . 98 : 40= 98g H2SO4
ms= md.100 : c
ms= 98 .100 : 20= 490 g sol. H2SO4
y=40 . 136 : 40 = 136 g CaSO4
n= 136g : 136g/moli = 1mol CaSO4
z= 40 . 2 : 40 = 2g H
n= 2g : 2g/moli = 1mol H2
VH2 = 1moli. 22,4 L/moli = 22,4 L
MCaSO4= 40 + 32 + 4.16= 136----> 136g/moli
MH2SO4= 2 + 32 + 4.16=98-----> 98g/moli
