Question

Two metallic oxides contain 27.6% and 30% oxygen respectively. If the formula of first oxide is $M_3{O}_4$, that of second will be:

• A. $MO$
• B. $M{O}_2$
• C. $M_2{O}_5$
• D. $M_2{O}_3$

Answer 1

The correct option is D $M_2{O}_3$

Given that, the formula of first oxide $=M_{3}O4$

Let mass of the metal = x

% of metal in $M_3{O}_4=(3x/3x+64)\times 100$

But as given percentage = (100-27.6) = 72.4 %

so, $(3x/3x+64)\times 100=72.4$

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 $M_2{O}_3$

Hence, the correct option is $\left(D\right)$.