mol HCl = 2 M×0.3 L= 0.6 mol setara dengan mol oksida logam
Bobot oksida logam =16 gram misal L = x gram, Mr= 56 dan O = (16-x ) gram, Mr =16 mol = gram/Mr Maka, g/Mr L + g/Mr O = mol yang setara dengan HCl.
x/56 + (16-x)/16 = 0.6 x/56 + 16/16 – x/16 = 0.6 (16x+896-56x)/896= 0.6 16x + 896-56x =537.6 358.4 = 40x x = 8.96 L= 8.96 gram O = 16-8.96= 7.04 gram
L : O = 8.96/56 : 7.04/16 = 0.2 : 0.4 = 1 : 2
Jadi rumus = LO2 L= Fe ( Mr= 56) Jadi rumus senyawa = FeO2 = besi (II ) oksida
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mol HCl = 2 M×0.3 L= 0.6 mol setara dengan mol oksida logam
Bobot oksida logam =16 gram
misal L = x gram, Mr= 56
dan O = (16-x ) gram, Mr =16
mol = gram/Mr
Maka,
g/Mr L + g/Mr O = mol yang setara dengan HCl.
x/56 + (16-x)/16 = 0.6
x/56 + 16/16 – x/16 = 0.6
(16x+896-56x)/896= 0.6
16x + 896-56x =537.6
358.4 = 40x
x = 8.96
L= 8.96 gram
O = 16-8.96= 7.04 gram
L : O = 8.96/56 : 7.04/16
= 0.2 : 0.4
= 1 : 2
Jadi rumus = LO2
L= Fe ( Mr= 56)
Jadi rumus senyawa = FeO2 = besi (II ) oksida