Balancing Silver(I)Nitrate & Potassium Iodide Reactions

In summary: because the first equation omits the potassium cations and the second equation has both the potassium and the nitrate anions.
  • #1
Clueless333
1
0
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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  • #2
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?

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

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

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

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

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

Remember those common solubility rules. They can help greatly :smile:
 
Last edited:
  • #3
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)
 

What is the purpose of balancing silver(I) nitrate and potassium iodide reactions?

The purpose of balancing these reactions is to ensure that the number of atoms of each element present on the reactant side is equal to the number of atoms on the product side. This helps us to accurately determine the amount of each substance needed for the reaction to occur and to predict the products.

Why is it important to balance these reactions?

Balancing reactions is important because it follows the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Therefore, the number of atoms of each element must be equal on both sides of the reaction in order for the reaction to be valid.

What are the steps for balancing a silver(I) nitrate and potassium iodide reaction?

The steps for balancing these reactions are as follows:

  • Write the unbalanced equation for the reaction.
  • Count the number of atoms of each element on both sides of the equation.
  • Add coefficients to the compounds to balance the number of atoms for each element.
  • Double-check to make sure the number of atoms of each element is equal on both sides.
  • Balance the charges if necessary by adding ions or changing the coefficients.
  • Write the final balanced equation.

Are there any tips for balancing these reactions more easily?

One tip for balancing these reactions is to start by balancing the elements that appear in only one compound on each side. Then, balance the elements that appear in two compounds on each side. Finally, balance the elements that appear in three or more compounds on each side.

What are some common mistakes to avoid when balancing these reactions?

Some common mistakes to avoid when balancing these reactions include:

  • Forgetting to check the number of atoms of each element after adding coefficients.
  • Not balancing the charges of ions if necessary.
  • Using fractions as coefficients instead of multiplying by the lowest common multiple.
  • Forgetting to include coefficients for compounds that have only one atom of an element.
  • Not double-checking the final balanced equation for accuracy.

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