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Question

25 g of CaCO3 completely reacts with H2SO4 and gives 8 g of CO2 then calculate the percentage purity of CaCO3.
CaCO3+H2SO4CaSO4+CO2+H2O

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Solution

Solution:-
Molecular weight of CaCO3=100g
Molecular weight of CO2=44g
Decomposition of CaCO3-
CaCO3CaO+CO2
Now, from the above reaction-
Weight of pure CaCO3 required to produce 44g of CO2=100g
Weight of pure CaCO3 required to produce 8g of CO2=10044×818.18g
Given weight of CaCO3=25g
Now,
Amount of pure CaCO3 in 25g of given MgCO3=18.18g
Thus,
Amount of pure CaCO3 in 100g of given MgCO3=18.1825×100=72.72g
Therefore,
The percentage purity of given CaCO3=72.72%

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