## M.9 hw prob

Doris Cho 1D
Posts: 53
Joined: Thu Jul 11, 2019 12:16 am

### M.9 hw prob

I've worked out most of the problem, but when I look at the solutions manual, I'm confused where they got the mole ratio of 2 mol NaOH over 1 mol Cu(NO3)2?
so when you're finding them oles of NaOH required to react with .04 mol Cu(NO3)2 in part B, how do you know there's 2 mols of NaOH?

also is something like part A likely to be on the test? because I didn't really understand how they got that...

NRodgers_1C
Posts: 54
Joined: Thu Jul 25, 2019 12:15 am

### Re: M.9 hw prob

I'm also having trouble seeing how the chemical equation from part a. relates to the equation from part b. I don't understand how to write "the net ionic equation for the reaction."

Natalie Nartz 4F
Posts: 50
Joined: Thu Jul 11, 2019 12:17 am

### Re: M.9 hw prob

The net ionic equation would look like this:

Cu+2 + 2OH- = Cu(OH)2

I hope this helps

Emma Popescu 1L
Posts: 105
Joined: Wed Sep 11, 2019 12:16 am

### Re: M.9 hw prob

The net ionic equation only lists the species participating in the reaction. We should learn it later in the course so don't worry too much about it for now.

Posts: 113
Joined: Thu Jul 11, 2019 12:16 am

### Re: M.9 hw prob

Natalie Nartz 4F wrote:The net ionic equation would look like this:

Cu+2 + 2OH- = Cu(OH)2

I hope this helps

Just a bit of clarification that I thought might be useful - the OH comes from NaOH; the Cu2+ comes from the copper (II) nitrate. Therefore, if you have 2 moles of OH- in your net ionic equation, you will have 2 moles of NaOH in your molecular equation. Hope this helps everything make a bit more sense!