Answer to Question #23302 in Chemistry for Taylor Bell

Question

how many grams of Cl are in 235 grams of CaCl2

Answer 1

Reaction of dissociation:

СaCl₂ <=> Ca²⁺ + 2Cl^-

The amount of substance for CaCl₂:

n(CaCl₂) = m(CaCl₂)/M(CaCl₂)
m(CaCl₂) = 235g
M(CaCl₂) = 40 + (35.5*2) = 111 g/mol
n(CaCl₂) = 235g/111g/mol = 2.18 mol
2n(Cl^-) = n(CaCl₂)
n(Cl^-) = 2.18*2 = 4.36 mol
m(Cl^-) = n(Cl^-)*M(Cl^-)
m(Cl^-) = 4.36 mol * 35.5 g/mol = 154.78 g

Answer:
m(Cl^-) = 154.78 g