Complete and Net Ionic Equations

Click For Summary

Discussion Overview

The discussion revolves around the process of writing molecular, complete ionic, and net ionic equations for a chemical reaction involving chromium(III) iodide and lead(II) nitrate in aqueous solution. The focus is primarily on balancing these equations correctly.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents their attempt at writing the molecular equation, complete ionic equation, and net ionic equation, expressing confusion about balancing the complete and net ionic equations.
  • Another participant agrees with the reduced net ionic equation provided, stating it accurately represents the formation of lead(II) iodide from its ions.
  • A third participant offers a formatting tip for displaying chemical equations, suggesting the use of specific tags for subscripts and superscripts.
  • The original poster expresses gratitude for the feedback received on their work.

Areas of Agreement / Disagreement

There appears to be agreement on the correctness of the reduced net ionic equation, but the original poster expresses uncertainty about their overall approach to balancing the equations. The discussion does not resolve the original poster's confusion.

Contextual Notes

The original poster's equations may depend on specific assumptions about the solubility of the compounds involved and the balancing process, which are not fully detailed in the discussion.

maceng7
Messages
18
Reaction score
0

Homework Statement


An aqueous solution of chromium(III) iodide is mixed with an aqueous solution of lead(II) nitrate

(a) Write the molecular equation for this reaction
(b) Write a complete ionic equation for this reaction
(c)If this reaction occurs, write the net ionic equation

I'm kinda lost on how to balance the complete and net ionic equations, this is what I've done so far:

The Attempt at a Solution



(a) 2CrI3 (aq) + 3Pb(NO3)2 (aq) -----> 2Cr(NO3)3 (aq) + 3PbI2 (s)

(b) 2(Cr)3+ (aq) + 6(I)- (aq) + 3(Pb)2+ (aq) + 6(NO3)- -----> 3PbI2 (s) + 2(Cr)3+ (aq) + 6(NO3)- (aq)

(c) 3(Pb)2+ (aq) + 6(I)- (aq) -----> 3PbI2 (s)
and if i divide all the coefficients by 3 then the cofficients are 1,2,1

I'm just sort of lost on writing the complete balance ionic equation. I would greatly appreciate someone looking over my work and tell me where I've gone wrong, thanks in advance
 
Physics news on Phys.org
I don't see anything wrong. The reduced equation in c) is exactly the normal production of lead(II) iodide from ions, so it is a good thing to reduce to.
 
Please note: you can format your equations using subscripts and superscripts. You can use X2 and X2 icons in the advanced editor, or [noparse] and [/noparse] tags.

For example

[noparse]NO3-[/noparse]

is displayed as

NO3-.
 
Thanks a lot guys, Borek I'll keep that in mind in the future thank you and Dchair thanks for looking over my work
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
1
Views
12K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K