E.29 part b HELP

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Paula Dowdell 1F
Posts: 72
Joined: Tue Nov 15, 2016 3:00 am

E.29 part b HELP

Postby Paula Dowdell 1F » 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!

Jennie Fox 1D
Posts: 66
Joined: Sat Jul 22, 2017 3:01 am

Re: E.29 part b HELP

Postby Jennie Fox 1D » 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-

Wenxin Fan 1J
Posts: 53
Joined: Thu Jul 13, 2017 3:00 am

Re: E.29 part b HELP

Postby Wenxin Fan 1J » 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

Katie Lam 1B
Posts: 52
Joined: Fri Sep 29, 2017 7:06 am

Re: E.29 part b HELP

Postby Katie Lam 1B » 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.


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