How many grams of 80% pure marble on calcination can give 14 grams of quick lime?

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25 gm

20 gm

31.25 gm

11.2 gm

answer is C.

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Detailed Solution

CaCO3→CaO+CO2 1 mole of CaO=1 mole of CaCO3 56 grams of CaO given by 100grams of CaCO3 14 grams of CaO given by (14/56)100=25 grams of CaCO3 As purity is 80% i.e. 80 grams of calcium carbonate is present in 100 grams of marble stone. 25 grams of calcium carbonate is present in (25 x 100 / 80) = 31.25 grams of marble stone

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