Question

2.6 gm of mixture of calcium carbonate and magnesium carbonate is strongly heated to constant weight of 1.3 gm. What is the weight of calcium carbonate in the original mixture?

Answer 1

Let the weight of CaCO₃ in mixture = x g
​Then weight of MgCO₃ will be 2.6-x g

​The CaCO₃ on heating produce CaO and CO₂ as follows:
CaCO₃ $\to$CaO + CO₂
​1 mole of CaCO₃ produce 1 mole of CaO.
Or, 100 g CaCO₃ produce 56 g CaO
therefore x g CaCO₃ will produce 0.56 x g. CaCO₃

​MgCO₃ on heating produce MgO and CO₂ as follows:
MgCO₃ $\to$MgO + CO₂
​​1 mole of MgCO₃ produce 1 mole of MgO.
Or, 84 g MgCO₃ produce 40 g MgO
therefore (2.6-x) g MgCO₃ will produce $\frac{40}{84}$(2.6-x) g MgCO₃

Given that fixed mass produced is 1.3 g
0.56 x + ​$\frac{40}{84}$(2.6-x) = 1.3 g
Multiplyied the eq. by 84 we get:
47.04 x + 104 - 40 x = 109.2 g
7.04 x = 5.2
x = 0.74 g

Weight of CaCO₃ in original mixture was 0.74 g