1.84g mixture of CaCO3 and MgCO3 was heated to a constant weight till 0.96 g residue formed % MgCO3 in sample was (1)45.66%(2) 54.34%(3)30%(4) 70%

Let the mass of CaCO3 be X gramSo,the mass of MgCo3 will be 1.84-X grams.Now,X grams of CaCO3 will give Cao=56X/100 gAnd,1.84-X grams of MgCO3 will give MgO=40(1.84-X)/84 gTotal mass of residue is 0.96 gSo,o.56+40(1.84-X)/84 g=0.96 g.°. X=1gramHence, % composition of CaCO3 =1/1.84×100=54.35%And,% composition of MgCO3=100%-54.35%=45.65%

Dear Student,
Please find below the solution to your problem.

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 + 84.3Y= 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=
= 0.98 / (0.98 + 0.86) × 100 = 53.26%

Mass % of MgCO3 in mixture =
100-53.26= 46.74%

Thanks and Regards