A sample of CaCO₃ and MgCO₃ weighed 2.21 g is ignited to constant weight of 1.152 g.The composition of mixture is :(Given molar mass in g mol⁻¹ : CaCO₃ : 100, MgCO₃ : 84)

Composition of CaCO₃–MgCO₃ Mixture | JEE Mains PYQ | Mole Concept

Q. A sample of CaCO₃ and MgCO₃ weighing 2.21 g is ignited to constant weight of 1.152 g. The composition of the mixture is :
(Given molar mass in g mol⁻¹: CaCO₃ = 100, MgCO₃ = 84)

Explanation

Step 1: Write decomposition reactions (on ignition)

CaCO₃ → CaO + CO₂
MgCO₃ → MgO + CO₂

Step 2: Assume masses of carbonates

Let mass of CaCO₃ = x g
Mass of MgCO₃ = y g
x + y = 2.21 ...(1)

Step 3: Use residue (oxide) masses

100 g CaCO₃ → 56 g CaO
84 g MgCO₃ → 40 g MgO

Mass of CaO formed = (56/100) x = 0.56x
Mass of MgO formed = (40/84) y

Total residue mass = 1.152 g
0.56x + (40/84)y = 1.152 ...(2)

Step 4: Solve equations (1) and (2)

From calculation:
x = 1.187 g
y = 1.023 g

Correct Answer = 1.187 g CaCO₃ + 1.023 g MgCO₃

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