Question

when the mixture of mgco3 and caco3 was heated for a long time,the weight decrease by 50% . calculate the % composition of the mixture.

Answer 1

MgCO3(s)→MgO(s) + CO2(g)

CaCO3(s)→CaO(s) + CO2(g)

The total initial weight of solid phase = (84x+100y) gm

x=the no. of moles of MgCO3 present

y=the no. of moles of CaCO3 present.

[MgCO3] = [MgO]

[CaCO3] = [CaO]

Final weight of solid phase = (40x + 56y) gm.

Final weight = 50% of the initial weight.

So, (84x + 100y)/2 = 40x + 56y

Or, 42x + 50y = 40x + 56y

Or, 2x = 6y

Or, x/y = 6/2 = 3

So, x:y = 3:1.

i.e 3 moles of MgCO3 =3×84=252 g

And 1 mole of CaCO3=100 g

In (252+100=352g) mixture.

So

% MgCO3=252×100/352=71.59% Ans

% CaCO3=100×100/352=28.41% Ans

__Er K100