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
B
M2O
C
MO
D
MO2

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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