I am having a bit of difficulty getting the net equation and the net ionic equation and the net ionic equation. I think I have the balance equation right. Thanks in advance for any help.
For the acid base neutralization reaction of carbonic acid, H2CO3(aq), and lithium hydroxide, LiOH(aq), write the balance:
a.)Complete equation
H2CO3 (aq) +2LiOH(aq) --> Li2CO3(aq) + 2H2O(l)
b.)Ionic equation
c.)Net ionic equation
H2CO3 (aq) +2LiOH(aq) --> Li2CO3(aq) + 2H2O(l)
Li2CO3 is only slightly soluble in aq solution; therefore, I would write (s) after Li2CO3
Ionic equation:
H2CO3(aq) + 2Li^+ + 2OH^- ==>Li2CO3(s) + H2O(l).
I would leave that as is for the net ionic equation unless you see something common to both sides of the equation.
im almost done with my hw just last six i don't get, i need the molecular, total ionic, and net ionic equations please
Aqueous solutions of potassium phosphate and calcium nitrate are mixed
.Aqueous solutions of barium chloride and sodium sulfate are mixed.
Magnesium is added to an aqueous solution of chromium (III) iodide.
Silver hydroxide is added to an aqueous phosphoric acid solution
.Aqueous solutions of sodium fluoride and hydrochloric acid are mixed.
Aqueous solutions of potassium dichromate and ammonium cyanide are mixed.
thank u so much for those who helped
To write the net ionic equation, we need to first write the balanced ionic equation.
a.) Complete equation:
H2CO3 (aq) + 2LiOH (aq) --> Li2CO3 (aq) + 2H2O (l)
b.) Ionic equation:
H2CO3 (aq) + 2Li+ (aq) + 2OH- (aq) --> 2Li+ (aq) + CO3^2- (aq) + 2H2O (l)
c.) Net ionic equation:
H2CO3 (aq) + 2OH- (aq) --> CO3^2- (aq) + 2H2O (l)
In the net ionic equation, we omit the spectator ions (ions that are present on both sides of the equation without undergoing any change). In this case, the spectator ions are Li+ and 2H2O.
Therefore, the net ionic equation for the acid-base neutralization reaction of carbonic acid and lithium hydroxide is:
H2CO3 (aq) + 2OH- (aq) --> CO3^2- (aq) + 2H2O (l)
To write the ionic equation and net ionic equation for an acid-base neutralization reaction, you need to first understand the reactants and products involved.
In this case, the reactants are carbonic acid (H2CO3) and lithium hydroxide (LiOH), and the products are lithium carbonate (Li2CO3) and water (H2O).
Let's break down the steps to write the ionic equation and net ionic equation:
a.) Complete equation:
First, you wrote the balanced chemical equation correctly:
H2CO3 (aq) + 2LiOH(aq) --> Li2CO3(aq) + 2H2O(l)
This equation represents the overall reaction in which the reactants combine to form products. It shows the balanced formula of all compounds involved.
b.) Ionic equation:
To write the ionic equation, you need to break down the soluble compounds into their respective ions. In this case, LiOH is a strong base that dissociates completely in water, as well as carbonic acid, which is a weak acid, but we will consider it as fully dissociated for simplicity.
H2CO3 (aq) + 2LiOH(aq) --> Li2CO3(aq) + 2H2O(l)
Ionizing the compounds gives:
2H+(aq) + CO3^2-(aq) + 2Li+(aq) + 2OH^-(aq) --> 2Li+(aq) + CO3^2-(aq) + 2H2O(l)
Note that the carbonate ion (CO3^2-) remains unchanged on both sides of the equation.
c.) Net ionic equation:
The net ionic equation is obtained by removing the spectator ions, which are the ions that appear on both sides of the equation and do not participate in the actual chemical reaction.
In this case, the spectator ions are the carbonate ions (CO3^2-). So, we can eliminate them from the equation to obtain the net ionic equation:
2H+(aq) + 2OH^-(aq) --> 2H2O(l)
The net ionic equation represents only the ions participating in the reaction and omits the spectator ions.
To summarize:
a.) Complete equation:
H2CO3 (aq) + 2LiOH(aq) --> Li2CO3(aq) + 2H2O(l)
b.) Ionic equation:
2H+(aq) + CO3^2-(aq) + 2Li+(aq) + 2OH^-(aq) --> 2Li+(aq) + CO3^2-(aq) + 2H2O(l)
c.) Net ionic equation:
2H+(aq) + 2OH^-(aq) --> 2H2O(l)