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Question

A sample of CaCO3 is 50% pure . On heating 1.12L of CO2 (at STP) is obtained . Residue left (assuming non-volatile impurity) is

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Solution

CaCO3 CaO +CO2100 g of CaCO3 gives 1 mole of carbondioxide = 22.4 LThat is 22.4 L of CO2 is obtained from 100 g of CaCO3So 1.12 L of CO2 is obtained from = 10022.4 × 1.12 = 5 g of CaCO3If the sample is 100 %purity the amount of CaCO3 = 5 gFor 50 % sample the amount of CaCO3 = 5100 × 50 = 2.5 gThe amount of residue left is calculated as follows100 g of CaCO3 gives 56 g of CaO as nonvolatile impurity2.5 g of CaCO3 gives = 56100 × 2.5 = 1.4 g of CaOThus the residue left = 1.4 g of CaO

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