Question

On reduction with hydrogen, $3.6$g of an oxide of metal left $3.2$g of metal. If the vapour density metal is $32$, the simplest formula of the oxide would be:

• A. $MO$
• B. $M_2{O}_3$
• C. $M_{2}O$
• D. $M_2{O}_5$

Answer 1

The correct option is C $M_{2}O$

Vapour density$=\frac{MolecularMass}{2}$

Molecular mass of metal $=2\times 32=64$

given

mass of metal$=3.2g$

no.of moles of metal$=\frac{3.2}{64}=\frac{1}{20}$

mass of oxygen$=3.6g-3.2g=0.4g$

no.of moles of oxygen$=\frac{3.2}{64}=\frac{1}{40}$

The Metal oxide be in simplest form

$M\frac{1}{20}O\frac{1}{40}\Rightarrow M_{2}O$