Home
Class 12
CHEMISTRY
On reduction with hydrogen, 3.6 g of an ...

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 is 64. the formula of the oxide is

A

`M_(2) O_(3)`

B

`M_(2) O`

C

MO

D

`MO_(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`M_(x)underset(3.6 g)( O_(y)) + H_(2) rarr underset(("Atomic weight "= 64))underset(3.2g)(xM) + y H_(2)O`
Moles of M = `(32)/(64) = (1)/(20) ` mol
Weight of oxygen = 36 - 32 = 0.4
Moles of oxygen = `(0.4)/(16) = (1)/(40)`
`( " moles of metal " )/( " moles of oxygen " ) = ((1)/(20))/( (1)/(4)) = 2 : 1`
`therefore " " M = 2 ` and
O = 1 or the formula of oxide is `M_(2) O`.
Hence, correct option is (b).
Promotional Banner

Similar Questions

Explore conceptually related problems

0.72 gm of an oxide of a metal M on reduction with H_(2) gave 0.64 g of the metal . The atomic weight of the metal is 64 . The empirical formula of the compound is

70 gms of a metal oxide on reduction produced 54 gms of metal. The atomic weight of the metal is 81. What is its valency?

What is the nature of metal oxides ?

The formula of a metal chloride is MCl_3 and it contains 20% of the metal.The atomic weight of the metal is approxiamately

When 6 xx 10^22 electrons are used in the electrolysis of a metallic salt, 1.9 gm of the metal is deposited at the cathode. The atomic weight of that metal is 57. So oxidation state of the metal in the salt is

Ametal oxide contains 53% of metal by weight. What is the equivalent weight of metal?