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Formula of first oxide = M3O4
let mass of the metal be == x
percentage of metal in M3O4 = (3x/ 3x+64) *100
but % age = (100-27.6) = 72.4 %
so, (3x/ 3x+64)*100 = 72.4
or x = 56.
in 2nd oxide,
oxygen = 30%....so metal = 70%
so, ratio :--
M : O
70/56 : 30/16
1.25 : 1.875
2 : 3
so, 2nd oxide = M2O3
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