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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80%

60%

90%

75%

answer is D.

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Detailed Solution

$\begin{array}{l} \underset{1 mole}{{CaCO}_{3} (s)} \to \underset{1 mole}{CaO (s)} + {CO}_{2} (g) \\ \underset{1 mole (56 g)}{CaO (s)} + 3 C (s) \to \underset{1 mole (64 g)}{{CaC}_{2} (s)} + CO (g) \\ \underset{64 g}{{CaC}_{2}} (s) + 2 H_{2} O \to \underset{26 g}{C_{2} H_{2}} (g) + Ca {(OH)}_{2} \\ If the yield of C_{2} H_{2} is 100 % then weight of \\ C_{2} H_{2} formed must be 26 g \\ but only 19 .5 g of C_{2} H_{2} is formced \\ ∴ % yiel of product = \frac{19 .5}{26} \times 100 = 75 % \end{array}$

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