Question

92 g mixture of CaCO3 , and MgCO3 heated strongly in an open vessel. After complete decomposition
of the carbonates it was found that the weight of residue left behind is 48 g. Find the mass of MgCO3
in grams in the mixture.​

Answer 1

Answer:

Explanation:

on heating it will split into oxides and co2

and co2 is liberated out

and only oxide weight is left

let weight of caco3 =x

weight of mgco3 = 92-x

weight of mgo =y

weight of  cao = 48-y

apply poac for ca

x/100 = 48-y / 56

56x= 480 - 100y

56x+100y=480

47.04x+84y=403.2

apply poac for  mg

92-x  / 84 = y / 40

3680 - 40x = 84 y

40x+84y = 3680

Answer 2

Given:

The mass of mixture of caCO3 and MgCO3 = 92 gm

The mass of residue left behind = 48 gm

To Find:

The mass of MgCO3  in grams in the mixture.​

Calculation:

- Let the mass of MgCO3 in the mixture = x gm.

⇒ The mass of CaCO3 in the mixture = (92 - x) gm.

- The thermal decomposition of CaCO3 and MgCO3 is given as:

MgCO3 → MgO + CO2

CaCO3 → CaO + CO2

- The residue given by 84 gm of MgCO3 = 40 gm

⇒ The residue given by x gm of MgCO3 = (40/84) × x gm

- The residue given by 100 gm of CaCO3 = 56 gm

⇒ The residue given by (92 - x) gm of CaCO3 = (56/100) × (92 - x) gm

- According to the question, we have:

The weight of residue = 48 gm

⇒ {(40/84) × x} + {(56/100) × (92 - x)} = 48

⇒ 0.476 x + 51.52 - 0.56 x = 48

⇒ 0.084 x = 3.52

⇒ x = 3.52/ 0.084

⇒ x = 41.9 gm

- So, the mass of MgCO3 in the mixture is 41.9 gm.