Question

Two metallic oxides contain $27.6$% 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 $X_2{O}_3$

Solution:

Given that, formula of first oxide = $M_3{O}_4$

Let mass of the metal = x

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

but as given percentage$=(100-27.6)=72.4$ $\left(\frac{3x}{3x+64}\right)\times 100=72.4$ or $x=56$

in 2nd oxide

$oxygen$ $=30$

so for the metal $=70$

so the ratio is,

$M:O=\frac{70}{56}:\frac{30}{161.25}:1.8752:3$

so the second oxide is,

$M_2{O}_3$

Hence the answer is

$M_2{O}_3$.($X_2{O}_3$)