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

Explanation

On calculation marble stone decomposes to give quick lime (CaO)
`CaCO_(3) to CaO+CO_(2)`
1 mole of CaO =1 mole of `CaCO_(3)`
56 grams of CaO =100 grams of `CaCO_(3)`
14 grams of CaO=?
The weight calcium carbonate required `=(14)/(56) xx 100`
=25 grams
Percentage purity is only 80%
80 grams of calcium carbonate =100 grams of marble stone 25 grams of calcium carbonate=?
Weight of marble stone to be calcinated `=(25)/(80) xx 100`
=31.25 grams