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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
M2O3
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B
M2O
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C
MO
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D
MO2
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Solution

The correct option is C $$
{\text{M}}_{\text{2}} {\text{O}}
$$
Let the metal oxide by $$M_xO_y$$.

$$M_xO_y+yH_2\longrightarrow xM+yH_2O$$
$$\Rightarrow $$ Moles of metal $$=\cfrac{3.2}{64}=0.05$$
Moles of metal oxides $$=\cfrac{3.6}{64x+16y}$$

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

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

$$\implies 64x+16y=72x$$

$$\implies 8x=16y$$

$$\implies x=2y$$

Simplest Formula is for $$x=2, y=1$$ 

The formula is $$M_2O$$.

Hence, option $$B$$ is correct.

Chemistry

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