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

Explanation

On calcination marble stone decomposes to give lime (CaO)
`CaCO_(3) rarr CaO+CO_(2)`
1 mol 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)xx100 = 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)xx100 = 31.25 grams`