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Sa se determine raportul de masa si compoziția procentuala pt urm oxizi: Co2, So3, MgO, Al2O3​

Răspuns :

CO2

r.a.=1: 2

r.m.=12: 32=3: 8

M=12+32=44------> 44G/MOLI

44g Co2-------12g C-------32 g O

100 g------------x--------------y

x=100.12 : 44=27,25%C

y=100 . 32 : 44=72,72% O

SO3

r.m.=32 : 48=4 : 6=2: 3

M=32 + 48=80-----> 80g/moli

80g SO3---------32 g S-------48 g O

100 g----------------x---------------y

x=100.32 : 80=

y=100.48 : 80-

MgO

r.m.=24 : 16=3: 2

M=24+ 16=40----->40g/moli

40g MgO----------24g Mg--------16g O

100g----------------x-------------------y

x=100.24 : 40=

y=100.16 : 40=

Al2O3

r.m.=2.27 : 3.16= 54 : 48=9: 8

M=54+ 48=102-----> 102g/moli

102 g Al2O3----------54gAl----------48 g O

100 g-----------------------x------------------y

x=100.54:102=

y=100 .48 : 102=