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 quick lime (CaO)
`CaCO_(3)rarrCaO+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)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" gram"`