One mole of calcium carbonate on thermal decomposition gives ‘X’ grams of residue. ‘X’ on heating with coke gets completely converted to ‘Y’ grams of solid. ‘Y’ grams of this solid on hydrolysis gives 19.5 grams of a gaseous product ‘Z’ . Percentage yeild of ‘Z’ obtained is
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answer is 4.
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Detailed Solution
CaCO3s1 mole→CaOs1 mole+CO2gCaOs1 mole56 g+3Cs→CaC2s1 mole64 g+COgCaC264 gs+2H2O→C2H226 gg+CaOH2If the yield of C2H2 is 100% then weight ofC2H2 formed must be 26 g but only 19.5g of C2H2 is formced∴%yiel of product=19.526×100=75%