Question

Two metallic oxides contains 27.6%and 30% oxygen

Answer 1

Answer:

Given that, formula of first oxide = M3O4

Let mass of the metal = x

% of metal in M3O4 = (3x/ 3x+64) *100

but as given % 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, the ratio is

M : O

70/56 : 30/16

1.25 : 1.875

2 : 3

so, 2nd oxide is M2O3

Hope this helps you

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