4 gram of an impure sample of CaCO3 on treatment with excess HCl produces 0.88 grams CO2. find percentage purity of CaCO3.
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CaCO3+2HCL→CaCl2+H2O+CO2
44g of CO2→100g of CaCO3
0.88g of CO2→2g of CaCO3
percentage of purity=2/4(100)=50%
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