Question

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Two oxide of metal contains 27.6% and 30% oxygen.

If the formula of first oxide is M3O4.

Find that of second ?


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Answer 1

Answer:

Let mass of the metal = x

% of metal in M

3

O

4

=(3x/3x+64)×100

But as given percentage = (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

Answer 2

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