prima metoda
Mg(OH)2
Mg(OH)2⇒Mg:O:H=24:32:2=12:16:1
29 partiMh(OH)2: contin 12 parti Mg, 16 parti O si 1 parte H
100 parti Mg(OH)2contin x parti Mg y parti O z parti H
x=100×12/29=41,37%Mg
y=100×16/29=55,17%O
z=100×1/29=3,44%H
Al2(SO4)3
Al:S:O= 54:96:192=9:16:32
57parti Al2(SO4)3 contine 9 parti Al ...16partiS.....32parti O
100parti Al2(SO4)3 x partiAl.......y partiS......zpartiO
x=100×9/57=15,78%Al
y=100×16/57=28,07%S
z=100×32/57=56,14%O
K2O
K:O=78:16=39:8
47parti K2O contin 39 parti K......8 parti O
100 partiK2O..............xparti k.......y partiO
x=100×39/47=82,97%K
y=100×8/47=17,02%O
a doua metoda
Mg(OH)2
MMg(OH)2=AMg+2AO+2AH=24+2×16+2×1=58g
58gMg(OH)2........24gMg.....32gO......2gH
100gMg(OH)2.......tgMg.........vgO........ugH
t=100×24/58=41,37%Mg
v=100×32/58=55,15%O
u=100×2/58=3,44%H
Al2(SO4)3
MAl2(SO4)3=2AAl+3AS+12AO=2×27+3×32+12×16=342g
342gAl2(SO4)3......2×27gAl......3×32gS......12×16gO
100gAl2(SO4)3.......xgAl............ygS.............zgO
x=100×2×27/342=15,78%Al
y100×2×32/342=9,35%S
z=100×12×16/342=56,14%O
K2O
MK2O=2AK+AO=2×39+16=94g
94gK2O....2×39gK... 16gO
100gK2O......xgK........ygO
x=100×2×39/94=82,97%K
y=100×16/94=17,02%O
