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### E.29 part b HELP

Posted: Wed Oct 04, 2017 11:54 pm
Hi I am having trouble with part b of E.29.
(a) 0.0417 mol CuCl2 x 4H2O
(b) The questions asks how many moles of Cl- ions are present in the sample. The book says to double the amount in part (a) but i don't really understand why. If it is asking specifically how many moles of Cl- ions are present why would I be doubling the moles of CuCl2 x 4H2O??
Thanks!

### Re: E.29 part b HELP

Posted: Thu Oct 05, 2017 12:04 am
Since you are solving for moles of Cl- ions, you have to convert the 0.0417 mol CuCl2 x 4H2O to moles of Cl-. Since one mole of CuCl2 x 4H2O contains 2 Cl- ions, you take the 0.0417 mol CuCl2 x 4H2O and multiply it by (2 mol Cl- / 1 mol CuCl2 x 4H2O).

This is what my work looks like:

0.0417 mol CuCl2x4H2O X (2 mol Cl-/1 mol CuCl2x4H2O) = 0.0834 mol Cl-

### Re: E.29 part b HELP

Posted: Thu Oct 05, 2017 1:31 am
There are 2 moles of Cl- for every mole of CuCl2. Therefore you just multiply the moles of CuCl2 from part A by 2. You yet 8.34x10^-2 moles Cl-

-Wenxin Fan 1H

### Re: E.29 part b HELP

Posted: Thu Oct 05, 2017 11:25 am
The number of moles of Cl- ions is different than the moles of CuCl2, so you use a stoichiometric ratio (2 mol Cl-/1 mol CuCl2) to get the correct moles of Cl- ions.