# F325 June 2011 Q8Watch

#1
The question is:

I can write the ionic equations for step 1. It is step 2 i'm having the problem with.

I know there is two ionic equations for step 2 and I can understand how to get the 2H+ + CO32- --> CO2 + H2O

However I don't understand how to work out the second ionic equation for step two.

This is what I've done so far:

The ionic equation for step one:

Cu + 4HNO3 ----> Cu2+ + 2NO3- + 2NO2 + 2H2O

I thought that the Cu2+ would react with the 2NO3- to give Cu(NO3)2

Therefore when Sodium Carbonate was added:

Cu(NO3)2 + Na2CO3 ----> CuCO3 + 2NaNO3

However when I deduce the ionic equation from this all of the ions cancel out.

Can you please show me where i'm going wrong? Thanks.
0
4 years ago
#2
CuCO3 is a solid, so Cu2+(aq) and CO32-(aq) are needed on the left.
0
4 years ago
#3
(Original post by _paul)
Cu + 4HNO3 ----> Cu2+ + 2NO3- + 2NO2 + 2H2O
It's not quite ionic yet, the nitric acid is dissociated in solution.

Can you please show me where i'm going wrong? Thanks.
This is much simpler than you're making it Step 2 is just:

CaCO3 + acid ----> CO2 + H2O + salt

It's not important what the acid is - just call it HA or something:

CaCO3 + HA ----> CO2 + H2O + CaA

Take it from there
4 years ago
#4
(Original post by EierVonSatan)
Step 2 is just:

CaCO3 + acid ----> CO2 + H2O + salt
That, Sir, is step 3.
0
4 years ago
#5
(Original post by Pigster)
That, Sir, is step 3.
Step 2 has two parts. But I just read that the OP was fine with the first part

drat.
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