Question

how to find percentage of calcium carbonate in limestone

Answer 1

Reaction:

CaCO3+2HCl⟶CaCl2+H2O+CO2CaCOX3+2HCl⟶CaClX2+HX2O+COX2

Number of milli equivalence of HClHCl used initially is 1515

Number of milli moles of HClHCl used initially is 152152

22 moles of HClHCl reacts with 11 mole of CaCO3CaCOX3.

So amount of calcium carbonate is 154×10−3 mol=38 g154×10−3 mol=38 g




We know that 2 moles of HClHCl are required to react completely with 1 mole of CaCO3CaCOX3. We find out the number of moles of HClHCl in 150ml150ml 0.1M0.1M HClHCl (0.1NHCl=0.1MlHCl0.1NHCl=0.1MlHCl), which is 0.0150.015moles. Now the amount of CaCO3CaCOX3reacted is (0.01520.0152) = 0.00750.0075 moles.

Amount of CaCO3CaCOX3 in gms =

0.0075 moles×100 (mol.wt of CaCO3)=0.75g0.0075 moles×100 (mol.wt of CaCO3)=0.75g

...implying the purity is 75

Answer 2

Answer:

The purity of $CaCO_3$ is 75%.

Explanation:

The chemical reaction taking place is :

$CaCO_3+2HCl$ ⟶ $CaCl_2+H_2O+CO_2$

Number of milli equivalence of $HCl$ used initially $=15$

Number of milli moles of $HCl$ used initially $=150$

$2$ moles of $HCl$ reacts with $1$ mole of $CaCO_3$.

We then find out the number of moles of $HCl$ in $150ml$, which is $0.015$ moles.

Now the amount of $CaCO_3$ reacted is $(\frac{0.015}{2} ) = 0.0075$ moles

Amount of $CaCO_3$ in gms = $0.0075$ moles * $100$ (mol.wt of $CaCO_3$ )$=0.75g$

Hence the purity of $CaCO_3$ is 75%.