Question

two oxides of metal contains 27.6% and 30% oxygen respectively if the formula of the first oxide is M3O4 then formula for second oxide is

Answer 1

Given that, formula of first oxide = M3O4
Let mass of the metal = x
% of metal in M3O4 = (3x/ 3x+64) x 100
but as given % age = (100-27.6) = 72.4 %
so, (3x/ 3x+64) x 100 = 72.4
or x = 56.
in 2nd oxide,
oxygen = 30%....so metal = 70%
so, the ratio is
M : O
70/56 : 30/16
1.25 : 1.875
2 : 3
so, 2nd oxide is M2O3

i hope it will help you
regards