calculate the numbers of mole of Cacl2 that can be obtained from 25g of limestone Caco3 in the presence (Ca:40 C:12 O:16 Cl35.5)
in the presence of what?
If CaCO3 is heated you can get
CaCO3 ==> CaO + CO2 so for every mole CaCO3 you start with you can get 1 mole of CaO and whatever that reacts with you can get 1 mol of CaCl2.
So mols CaCO3 in 25 g CaCO3 = 25/molar mass CaCO3 = 25/100 = 0.25. Therefore you can obtain 0.25 mol CaCl2.