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Offline jennielynn_1980

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net ionic equations
« on: July 10, 2006, 12:43:37 PM »
I have to write ionic equations for the following and I am unsure of how to do so.  Here is the question and my attempt at each:

a) chromium dipped into silver nitrate
Cr3+ +Ag+(NO3)- --> Cr 3+(NO3)- +Ag+

b) gold immersed in hydrocholoric acid
NO REACTION and therefore NO EQUATION

c) nickel pellets dropped into a bath of calcium acetate
Ni2+ + Ca2+ (C2H3O2)2- --> Ca2+ + Ni 2+ (C2H3O2)2-

d) aluminium dropped into a bath of sulphuric acid
Al3+ + H22+(SO4)2- --> H22+ + Al3+(SO4)2-

e) zinc dipped inot lead(II) nitrate

Zn2+ + Pb2+ (NO3-)2 --> Pb 2+ +Zn 2+ (NO3)2

I have the rules for writing the equations but the examples given are not helping me answer these questions.  I think my mistakes are:
1) I am only supposed to ionize (aq) substances and I ionized everything
2) when writing a compound I am supposed to just give the charge of the whole molecule instead of each part.

Other than that I am lost.

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Re: net ionic equations
« Reply #1 on: July 10, 2006, 01:45:35 PM »
1) yes, this is the problem. "Chromium" is Chromium metal, Cr NOT the ion, likewise Nickel is Ni, etc.

2) You can give the charge for each part, that's fine, and even write them seperately as this will make it easier to spot the spectator ions.

eg. Lithium dropped into aqueous HBr

HBr(aq) <-----> H+(aq) + Br-(aq)

Li(s) + H+(aq) + Br-(aq) -----> LiBr(aq) + 1/2H2(g)

But,

LiBr(aq) <-----> Li+(aq) + Br-(aq)

So overall,

Li(s) + H+(aq) + Br-(aq) -----> Li+(aq) + Br-(aq) + 1/2H2(g)

So Bromide is a spectator ion, and the net ionic equation is

Li(s) + H+(aq) -----> Li+(aq) + 1/2H2(g)

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Offline jennielynn_1980

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Re: net ionic equations
« Reply #2 on: July 10, 2006, 02:20:02 PM »
Okay so a) might look like this:

a) chromium dipped into silver nitrate
Cr(s) +Ag+(NO3)- (aq)--> Cr3+(NO3)- +Ag(s)

Then seperate the aqueous ions right? so:

Cr(s) + Ag+(aq) + NO3-(aq) --> Cr3+(aq) + NO3-(aq) + Ag(s)

Then we can see NO3 is a spectator ion so we would rewrite it as:

Cr(s) + Ag+(aq) --> Cr3+(aq) + Ag(s)

But somehow I think this is still wrong  ???  Also, I can't figure out how solubility plays in.  I am seriously confused and frustrated!
« Last Edit: July 10, 2006, 04:44:22 PM by jennielynn_1980 »

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Re: net ionic equations
« Reply #3 on: July 11, 2006, 06:52:01 AM »
Quote
Cr(s) +Ag+(NO3)- (aq)--> Cr3+(NO3)- +Ag(s
Your formula for Chromium(III) Nitrate is wrong and the charge is not balanced. It should be Cr(NO3)3

Quote
Cr(s) + Ag+(aq) --> Cr3+(aq) + Ag(s)
Your previous mistake carries forward itself here. It resulted in your charges are not balanced. It should be:

Cr(s) + 3 Ag+(aq) --> Cr3+(aq) + Ag(s)
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline jennielynn_1980

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Re: net ionic equations
« Reply #4 on: July 11, 2006, 01:45:04 PM »
I figured out e) and c). 

I have a question about d.  If I end up with 6H (aq) on one side of the equation and 6H (g) on the other side, do I still cross it out as spectator ions?


For a) I am still trying to balance it out after correcting my mistake in the formula for chromium(III)nitrate.  Here is what I have with the revisions:

chromium + silver nitrate --> silver + chromium nitrate

3Cr(s) + 3Ag(NO3)(aq) --> 3Ag(s) + Cr(NO3)3 (aq)

3Cr(s) + 3Ag+ (aq) + 3NO3-(aq) --> 3Ag(s) + 3Cr3+ (aq) + 3NO3- (aq)

eliminate the NO3 as a spectator ion and we have:

3Cr(s) + 3Ag+ (aq) --> 3Ag(s) + 3Cr3+ (aq)

The equation is balanced but the charge is not.  What am I missing?

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Offline Dan

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Re: net ionic equations
« Reply #6 on: July 11, 2006, 02:42:18 PM »
I have a question about d.  If I end up with 6H (aq) on one side of the equation and 6H (g) on the other side, do I still cross it out as spectator ions?

Do you mean 6H+(aq) and 3H2(g)?

If the species does not remain exactly the same, you do not cross it out.

Quote
chromium + silver nitrate --> silver + chromium nitrate

3Cr(s) + 3Ag(NO3)(aq) --> 3Ag(s) + Cr(NO3)3 (aq)


Count your Cr, you have balanced it incorrectly, you have 3 on the left and one on the right.

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Offline jennielynn_1980

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Re: net ionic equations
« Reply #7 on: July 12, 2006, 11:50:47 AM »
Thanks :)

I did realize that it was balanced incorrectly eventually.  I also had an incorrect formula for one of the other questions.  Now that is sorted, I think I got them all.

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