Question

0.36 gm of a metal when burnt in oxygen yields 0.60 gm of oxide. The carbonate of that metal contains 28.57% of the metal. Assuming the validity of the law of definite proprotioons, determine the weight of the oxide formed by heating 1 gm of that carbonate.

Answer 1

Explanation:

0.36 gm of metal gives 0.6 gm of oxide, thus the amount of oxygen added is 0.6-0.36 = 0.24 g of O

ratio of reactivity = 0.36:0.24 = 3:2

Carbonate will have structure as MCO3 =

Wt. of CO3 = 12 + (16x3) = 60gms.

It is said that, MCO3 has 28.75% metal i.e. remaining 71.25% is CO3

i.e. from if 100 gms contains 71.25 gm of CO3 then taking ratios = 71.25/100 = 60/?

?= 84

i.e. 84 gm of carbonate will contain 60 gm of CO3 and 24 gm of metal.

Ration of reacivity = 24:60 i.e. 2:5

When MCO3 is heated it will form MO with liberation of CO2

i.e. 84 - wt. of CO2 = 84-44 = 40 gms of oxide

Ratio = 84:40 then 1:?

0.48 gms of oxide from 1 g of carbonate

Answer 2

0.36 gm of metal gives 0.6 gm of oxide, thus the amount of oxygen added is 0.6-0.36 = 0.24 g of O

ratio of reactivity = 0.36:0.24 = 3:2

Carbonate will have structure as MCO3 =

Wt. of CO3 = 12 + (16x3) = 60gms.

It is said that, MCO3 has 28.75% metal i.e. remaining 71.25% is CO3

i.e. from if 100 gms contains 71.25 gm of CO3 then taking ratios = 71.25/100 = 60/?

?= 84

i.e. 84 gm of carbonate will contain 60 gm of CO3 and 24 gm of metal.

Ration of reacivity = 24:60 i.e. 2:5

When MCO3 is heated it will form MO with liberation of CO2

i.e. 84 - wt. of CO2 = 84-44 = 40 gms of oxide

Ratio = 84:40 then 1:?

0.48 gms of oxide from 1 g of carbonate