In an experiment, 4 g of M₂O_x oxide was reduced to 2.8 g of the metal. If the atomic mass of the metal is 56 g mol^-, the number of O atoms in the oxide is

${\mathrm{M}}_2{\mathrm{O}}_{\mathrm{x}}\stackrel{\text{ Reduction }}{⟶}\mathrm{M}$
$\text{ Eq. of }{\mathrm{M}}_2{\mathrm{O}}_{\mathrm{x}}=\mathrm{eq}\text{. of Metal }$
$\frac{\text{ Wt. of }{\mathrm{M}}_2{\mathrm{O}}_{\mathrm{x}}}{\text{ Eq. wt. of }{\mathrm{M}}_2{\mathrm{O}}_{\mathrm{x}}}=\frac{\text{ Wt. of Metal }}{\text{ Eq. wt. of Metal }}$
$\frac{\frac{4}{2\times 56+\mathrm{x}\times 16}}{2\mathrm{x}}=\frac{2.8}{\frac{56}{\mathrm{x}}}$---------(i)
On solving we get,
$\frac{4}{56+8\mathrm{x}}=\frac{2.8}{56}\Rightarrow \frac{1}{14+2\mathrm{x}}=\frac{1}{20}\Rightarrow 2\mathrm{x}=6\Rightarrow \mathrm{x}=3$
Hence, the oxide is M₂O₃.