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Answer on Other Chemistry Question for Taylor Bell

Question #23302
how many grams of Cl are in 235 grams of CaCl2
Expert's answer
Reaction of dissociation:

–°aCl2 <=> 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

m(Cl-) = 154.78 g

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