Question

Two metallic oxides contains $27.6\%$ and $30\%$ oxygen respectively. If the formula of the first oxide is $X_3{O}_4$, that of the second will be:

• A. $XO$
• B. $X{O}_2$
• C. $X_2{O}_5$
• D. $X_2{O}_3$

Answer 1

The correct option is D $X_2{O}_3$

Given that, formula of first oxide $=X_3{O}_4$

Let mass of the metal $=x$

% of metal in $X_3{O}_4=\frac{3x}{3x+64}\times 100$

but as given % age $=(100-27.6)=72.4$ %

So, $\frac{3x}{3x+64}\times 100=72.4$

or $x=56.$

In 2nd oxide,

oxygen = 30%, so metal = 70%

So, the ratio is

=>$X:O$ =>$\frac{70}{56}:\frac{30}{16}$

=>$1.25:1.875$

=>$2:3$

So, 2nd oxide is $X_2{O}_3$