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The reaction:-
Molecular wt of CaCO3=40+12+(3x16)=100
So, gram molecular weight of CaCO3=100 gm
Molecular wt of CO2=12+(2x16)=44
So, gram molecular wt of CO2=44 gm
100 gm of CaCO3 gives 44 gm of CO2.
OR, 1 gm of CO2 is produced from (100/44) gm of CaCO3
OR, 3.52 gm of CO2 is produced from {(100x3.52)/44} gm = 8 gm of CaCO3
So, 8 gm CaCO3 present in 40 gm limestone.
So, % purity:-
(8x100)/40% = 20% purity
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