Respuesta :

First we determine the moles CaCl2 present:

525g / (110.9g/mole) = 4.73 moles CaCl2 present 

Based on stoichiometry, there are 2 moles of Cl for every mole of CaCl2:
(2moles Cl / 1mole CaCl2) x 4.73 moles CaCl2 = 9.47 moles Cl 

Get the mass:
9.47moles Cl x 35.45g/mole = 335.64 g Cl