Question

4 Two oxides of a metal contain 53.4% and 36.4% of
oxygen by mass respectively. If the formula of first oxide
is MO, then that of the second is
(2) Mo (6) M20 (C) MOS (6) MO2.
5 In an experiment 4 g of MO oxide was reduced to​

Answer 1

Answer:

The answer is M2O.

Explanation:

The ratio of metal and oxygen in the first oxide, M3O4 = 46.6: 53.4

The ratio of metal and oxygen in the second oxide =63.6: 36.4

Let the molecular mass of metal = M

Now calculate the percentage by weight of the metal in the oxide:

 M x 100 / M + 16 = 46.6

100M = 46.6M + 745.6

100M - 46.6M = 745.6

53.4M = 745.6

M = 745.6/53.4

M =13.96 = 14 g

Now,

Moles of metal in second oxide = 63.6 / 14 = 4.54

Moles of oxygen in second oxide = 36.4 / 16 = 2.275

Ration of moles of metal and oxygen = 4.54: 2.275

= 2 :1

Thus formula of secong oxide is M2O.

Answer 2

Answer:

M2O

Explanation:

53.4% O and 100-53.4= 46.6% M in first oxide.

let x be mass of metal,

x × 100/x+16= 46.6

by solving x= 14g

36.4% O and 100-36.4=63.6% M in second oxide.

by finding the number of moles of both oxygen and metal we get,

n of M= 63.6/14=4.5

n of O=36.4/16=2.2

by taking the ratio of number of moles,

the formula becomes M2O