On reduction with hydrogen, 3.6 ~g of an oxide of metal (M) left...

On reduction with hydrogen,

3.6 \mathrm{~g}

of an oxide of metal

(M)

left

3.2 \mathrm{~g}

of the metal. If the atomic weight of the metal is 64 . The formula of the oxide is On reduction with hydrogen, 3.6 ~g of an oxide of metal (M) left 3.2 ~g of the metal. If the atomic weight of the metal

On reduction with hydrogen, 3.6 ~g of an oxide of metal (M) left 3.2 ~g of the metal. If the atomic weight of the metal

Options

A $\mathrm{M}_2 \mathrm{O}_3$
B $\mathrm{M}_2 \mathrm{O}$
C $M O$
D $\mathrm{MO}_2$

Correct answer

B. $\mathrm{M}_2 \mathrm{O}$

Step-by-step solution

Moles of

M=\frac{3.2}{64}=\frac{1}{20} \mathrm{~mol}

Weight of oxygen

=3.6-3.2=0.4

Moles of oxygen

=\frac{0.4}{16}=\frac{1}{40}

\begin{aligned}& \frac{\text { Moles of metal }}{\text { Moles of oxygen }}=\frac{\frac{1}{20}}{\frac{1}{40}}=2: 1 \\ & \therefore \quad M=2 \text { and }\\ & \therefore \quad \mathrm{O}=1 \text { or the formula of oxide is }M_2 \mathrm{O} . \end{aligned}

$$Hence, correct option is (b).

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