Explain in detail....

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 .

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

On heating the mixture...

$C a C O_{3} + M g C O_{3} \overset{Δ}{\to} C a O + M g O + 2 {C O}_{2}$

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.

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

Solving equation 1 and 2 we get.

Y=0.01

X=0.009

Thus mass of CaCO₃in mixture =100 X 0.0098 =0.98 g

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

Mass % of CaCO₃in mixture=

$\frac{0.98}{1.84} \times 100 = 53.26 %$

Mass % of MgCO₃ in mixture =

100-53.26= 46.74%