Complete and Net Ionic Equations

[maceng7]

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

 

[Dchair]

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.

 

[Borek]

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^-.

 

[maceng7]

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

 

[DeeNos]

!
Your molecular equation looks correct, but your complete ionic equation is not balanced. Here is the correct complete ionic equation:

2Cr3+ (aq) + 6I- (aq) + 3Pb2+ (aq) + 6NO3- (aq) -----> 2Cr3+ (aq) + 6NO3- (aq) + 3PbI2 (s)

To balance the equation, you need to make sure that the number of each type of atom is the same on both sides of the equation. In this case, you can see that there are 6 nitrate ions on both sides, so that part is already balanced. To balance the chromium ions, you need to add a coefficient of 2 in front of the chromium ions on the right side. This will give you 4 chromium ions on both sides. Similarly, to balance the iodide ions, you need to add a coefficient of 3 in front of the iodide ions on the left side. This will give you 6 iodide ions on both sides. The final balanced complete ionic equation is:

2Cr3+ (aq) + 6I- (aq) + 3Pb2+ (aq) + 6NO3- (aq) -----> 2Cr3+ (aq) + 6NO3- (aq) + 3PbI2 (s)

To write the net ionic equation, you need to eliminate any spectator ions, which are ions that are present on both sides of the equation and do not participate in the reaction. In this case, the nitrate ions are spectator ions, so they can be eliminated. The final balanced net ionic equation is:

2Cr3+ (aq) + 6I- (aq) + 3Pb2+ (aq) -----> 3PbI2 (s)

I hope this helps clarify the process of balancing complete and net ionic equations. Remember to always check that the number of each type of atom is the same on both sides of the equation and to eliminate spectator ions when writing the net ionic equation.