67 252
Assignments Done
99,3%
Successfully Done
In November 2018

Answer to Question #23302 in Other Chemistry 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

Answer:
m(Cl-) = 154.78 g

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

Submit
Privacy policy Terms and Conditions