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)

Solution

Chemistry ma’am · 3 steps

Ignition drives off CO₂, so track what fraction of each carbonate stays behind as oxide.

1. What ignition does. Both carbonates lose CO₂: CaCO₃ → CaO + CO₂ and MgCO₃ → MgO + CO₂. So 100 g of CaCO₃ leaves 56 g of CaO, and 84 g of MgCO₃ leaves 40 g of MgO.
2. Two conditions. Let the masses be x g of CaCO₃ and y g of MgCO₃. The carbonates must add up to 2.21 g, and the oxides left behind must add up to 1.152 g.

   x + y = 2.21, 56/100x + 40/84y = 1.152

   Option	Total carbonate (g)	Oxide left, 0.56x + 0.476y (g)
   (a)	2.374, not 2.21	1.230
   (b)	1.187 + 1.023 = 2.21	0.665 + 0.487 = 1.152
   (c)	1.023 + 1.187 = 2.21	0.573 + 0.565 = 1.138
   (d)	2.046, not 2.21	1.060

3. So. Only (b) satisfies both conditions. The mixture has 1.187 g of CaCO₃ and 1.023 g of MgCO₃.

The idea to keep. For a mixture that is heated, write one equation for the starting mass and one for the residue, using the mass fraction of each compound that survives heating.

Answer from the NTA final answer key. Solution by PrepWiser.