Question

In an experiment, 4 g of a metallic oxide was reduced
to 2.8 g of the metal. If the atomic mass of the metal is
56 g mor. the formula of the oxide is
(a) M2O
(b) MO
(c) M2O3
(d) MO2
​

Answer 1

Answer

c) M₂O₃

Explanation

Given, 4 g of metallic oxide is reduced to 2.8 g of the metal.

⇒ amount of oxygen removed from the oxide to get the metal = 4g - 2.8g = 1.2 g

Which means, 2.8 g of metal combines with 1.2 g of oxygen to form metallic oxide.

Given molar mass of metal = 56 g

⇒ mole = given mass/molar mass = 2.8/56 = 1/20 mole

And, we know that molar mass of oxygen = 16 g

⇒ mole = 1.2/16 = 3/40 mole

This means. that 1/20 mole of metal combines with 3/40 mole of oxygen to form metal oxide.

So, 2 mole of metal would combine with 3 mole of oxygen to form metal oxide.

⇒2 atom of metal requires 3 atoms of oxygen to form metal oxide.

This gives the formula M₂O₃ for the metallic oxide.

Answer 2

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