Question

On reduction with hydrogen, 3.6 g of an oxide of metal left 3.2 g of the metal. If the atomic weight of the metal is 64, the simplest formula of the oxide would be:

• A. ${\text{M}}_{\text{2}}{\text{O}}_{\text{3}}$
• B. ${\text{M}}_{\text{2}}\text{O}$
• C. $\text{M}\text{O}$
• D. ${\text{MO}}_{\text{2}}$

Answer 1

Answer: (B)

${\text{M}}_{\text{2}}\text{O}$

Let the metal oxide by $M_x{O}_y$.

$M_x{O}_y+y{H}_2⟶xM+y{H}_{2}O$

$\Rightarrow$ Moles of metal $=\frac{3.2}{64}=0.05$

Moles of metal oxides $=\frac{3.6}{64x+16y}$

$1$ $mole$ of metal oxide gives $x$ $mole$ metal $M$

$\frac{0.05}{x}=\frac{3.6}{64x+16y}$

$⟹64x+16y=72x$

$⟹8x=16y$

$⟹x=2y$

Simplest Formula is for $x=2,y=1$

The formula is $M_{2}O$.

Hence, option $B$ is correct.