Balancing Silver(I)Nitrate & Potassium Iodide Reactions

AI Thread Summary
The discussion focuses on balancing chemical reactions and writing net ionic equations. For the reaction of calcium with water, the balanced equation is Ca + 2H2O → Ca(OH)2 + H2, with calcium hydroxide being soluble and thus not included in the net ionic equation. In the reaction between silver(I)nitrate and potassium iodide, the net ionic equation is Ag+ + I- → AgI(s), as silver iodide is insoluble while the other products are soluble. Participants emphasize the importance of recognizing soluble versus insoluble compounds when writing net ionic equations. Understanding solubility rules is crucial for accurately determining these equations.
Clueless333
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1. Calcium + Water yields calcuim hydroxide + hydrogen
Need balanced and net ionic equation, I got
Ca + 3/2H20 =Ca(OH)2 +H
I can't figure out how to balance it further and I don't know how to get the net equation cause calcuim hydroxide is soluble which means it wouldn't be in the net equation right?

2. Silver(I)nitrate + potassium iodide yields silver(I)iodide + potassium nitrate
I got
AgNO2- + KI = AgI + KNO2-
Ag+ + I- = AgI
Potassium nitrate is soluble so it's not in the net equation but silver iodide is not soluble so it is. Is it balanced ok? Am I right? Heeelllpp!
 
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Clueless333 said:
1. Calcium + Water yields calcuim hydroxide + hydrogen
Need balanced and net ionic equation, I got
Ca + 3/2H20 =Ca(OH)2 +H
I can't figure out how to balance it further and I don't know how to get the net equation cause calcuim hydroxide is soluble which means it wouldn't be in the net equation right?

\text{Ca}}\left( {{\text{OH}}} \right)_2 is soluble in water.
Therefore, you will observe hydrolysis, as:
{\text{H}}_{\text{2}} {\text{O}} \rightleftharpoons {\text{H}}^ + \left( {aq} \right) + {\text{OH}}^ - \left( {aq} \right)

*Or you balance it as:
2{\text{H}}_{\text{2}} {\text{O}} \rightleftharpoons {\text{H}}_{\text{3}} {\text{O}}^{\text{ + }} \left( {aq} \right) + {\text{OH}}^ - \left( {aq} \right)

Clueless333 said:
Silver(I)nitrate + potassium iodide yields silver(I)iodide + potassium nitrate .. ..
*Actually, silver(I) nitrate is--> \text{AgNO} _ 3 :wink:

You are right, \text{KNO} _ 3 and \text{AgNO} _ 3 and {\text{KI}} are all obviously soluble. Looking at your chemicals, you will observe {\text{AgI}} precipitate. Therefore, the net ionic reaction equation will omit the potassium cations and the nitrate anions:

{\text{Ag}}^ + \left( {aq} \right) + {\text{I}}^ - \left( {aq} \right) \rightleftharpoons {\text{AgI}}\left( s \right)

Remember those common solubility rules. They can help greatly :smile:
 
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ok, I understand how to work out soluble and nonsoluble compounds but I do not know how to write Net Ionic Equations. So if I had:

BaCl2(aq) + 2AgNO3(aq) -->Ba(NO3)2(aq)+2AgCl(s)

Would the Net Ionic equation be:
a) 2Ag+(aq) + 2Cl-(aq) -->2Ag+Cl-(s)

or
b) Ag+(aq) + Cl-(aq) --> Ag+Cl-(s)

or
c) 2Ag+(aq) + Cl2-(aq) --> 2Ag+Cl-(s)

My current guess would be (c)
 
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