when the mixture of MgCO3 and CaCO3 was heated for a long time. The we

1.	when the mixture of MgCO3 and CaCO3 was heated for a long time. The weight decreased by 50%. Findthe percentage composition of the mixture.
Answer» 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

when the mixture of MgCO3 and CaCO3 was heated for a long time. The weight decreased by 50%. Findthe percentage composition of the mixture.

Answer»

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.

% MgCO3=252×100/352=71.59% Ans

% CaCO3=100×100/352=28.41% Ans