Определить массовую долю: cr2o3 и znco3 заранее

M(Cr2O3)=52*2+16*3= 152 g/mol
W(Cr)=Ar(Cr)*n / M(Cr2O3)*100% = 52*2 / 152*100%=68.42%
W(O) = 100%-W(Cr)=100-68.42=31.58%

M(ZnCO3)=65+12+16*3=125g/mol
W(Zn)=65/125*100%=52 %
W(C)=12/125*100% = 9.6%
W(O)=16*3 / 125 *100%=38.4 %

Mr(Cr2O3) = 52*2 + 16*3 = 152
W(Cr) = Ar(Cr) *n / Mr(Cr2O3)  *100% = 52*2 / 152  *100% =68%
W(O) = Ar(O) *n / Mr(Cr2O3)  *100% = 16*3 / 152  *100% = 32%

Mr(ZnCO3) = 65 + 12 + 16*3 = 125
W(Zn) = Ar(Zn) *n / Mr(ZnCO3)  *100% = 65*1 / 125  *100% = 52%
W(C) = Ar(C) *n / Mr(ZnCO3)  *100% = 12 / 125  *100% = 9,6%
W(O) = Ar(O) *n / Mr(ZnCO3)  *100% = 16*3 / 125  *100% =38,4%