Question

A sample of CaCo3 and MgCo3 weighing 192 g is ignited to constant weight of 104 g What is the composition of mixture?

Answer 1

Molar mass of CaCO3 = 100g/mol

Molar mass of MgCO3=84.3 g/mol

Let number of moles of CaCO3 and MgCO3 in the mixture is X & Y respectively .

So

100X+84.3 Y= 1.84.......................(1)

 

On heating the mixture...

CO2 goes off from the product but mixture of  CaO and MgO remains in the solid form.

Molar mass of CaO= 56g/mole

Molar mass of MgO= 40.3 g/mole

From the equation it is clear that number of moles of  MgO & CaO in product mixture is same as that of respective carbonates in the reactant mixture.

so

56X+40.3Y=0.96........................(2)

Solving equation 1 and 2 we get.

 

Y=0.01

X=0.009

 

Thus mass of CaCO3 in mixture =100 X 0.0098 =0.98 g

Mass of  MgCO3 in mixture = 84.3 X 0.01= 0.86 g

 Mass % of CaCO3 in mixture=

Mass % of  MgCO3 in mixture =

100-53.26= 46.74%

Answer 2

Answer:

CaCO3 ------ CaO + CO2

MgCO3 ------ MgO +CO2

let mass of CaCO3 be x gm, moles = x/100

so mass of MgCO3 is (192 - x)gm, moles = (192-x)/84

mass of CaO = x/100 * 36 = 36x/100 gm

mass of MgO = (192-x)/84 * 40 = (1920-10x)/21 gm

Now 36x/100 + (1920-10x)/21 = 104  

find the value of x by this  

you will get mass of CaCO​3

then find the composition