What is the total ionic equation of this:
3CaCl2(aq) + 2K3PO4(aq) → Ca3(PO4)2(s) + 6KCl(aq)
Shw all physical states of the reactants and products
John, Mandi, whomever--
You should go back and look at your other posts. I answered this below an hour ago. Here is a link.
http://www.jiskha.com/display.cgi?id=1411784311
Well, buckle up for a chemistry comedy show! Here's the total ionic equation and the physical states of the reactants and products:
3Ca²⁺(aq) + 6Cl⁻(aq) + 6K⁺(aq) + 2PO₄³⁻(aq) → Ca₃(PO₄)₂(s) + 6K⁺(aq) + 6Cl⁻(aq)
And now for our star-studded cast:
- "Ca²⁺(aq)" and "Cl⁻(aq)" are swimming in the aqueous solution.
- "K⁺(aq)" is also rocking the aqueous party.
- "PO₄³⁻(aq)" is ready to mingle in the aqueous solution as well.
- "Ca₃(PO₄)₂(s)" is the grand finale; it's gonna be a solid show.
- And last but not least, "K⁺(aq)" and "Cl⁻(aq)" are jumping in as co-stars in the aqueous party again.
And that's a wrap, folks! Chemistry never fails to mix a little drama with all those elements. Stay tuned for more laughs and learning!
To write the total ionic equation, we need to separate all the species into their respective ions.
The given chemical equation is:
3CaCl2(aq) + 2K3PO4(aq) → Ca3(PO4)2(s) + 6KCl(aq)
Here are the ions for each compound:
3CaCl2(aq) can be written as:
3 Ca^2+(aq) + 6 Cl^-(aq)
2K3PO4(aq) can be written as:
6 K^+(aq) + 2 PO4^3-(aq)
Ca3(PO4)2(s) remains unchanged as it is a solid.
6KCl(aq) can be written as:
6 K^+(aq) + 6 Cl^-(aq)
Now that all the species have been written in their ionic forms, we can write the total ionic equation:
3 Ca^2+(aq) + 6 Cl^-(aq) + 6 K^+(aq) + 2 PO4^3-(aq) → Ca3(PO4)2(s) + 6 K^+(aq) + 6 Cl^-(aq)
Remember, the solid Ca3(PO4)2 does not break apart into ions, so it remains unchanged.
Finally, to simplify the equation, we can cancel out the ions that appear on both sides of the equation:
3 Ca^2+(aq) + 2 PO4^3-(aq) → Ca3(PO4)2(s)
This is the total ionic equation for the given reaction.
To write the total ionic equation of the given reaction, we need to break down all the ionic compounds into their individual ions (called dissociation) and indicate their appropriate charges. Here's how you can do it:
First, write out the balanced molecular equation:
3CaCl2(aq) + 2K3PO4(aq) → Ca3(PO4)2(s) + 6KCl(aq)
Now, let's break down each compound into its constituent ions, indicating the physical states:
Starting with CaCl2(aq), it dissociates into Ca2+ and 2Cl- ions:
3CaCl2(aq) → 3Ca2+(aq) + 6Cl-(aq)
Next, we have K3PO4(aq), which dissociates into 3K+ and PO4^3- ions:
2K3PO4(aq) → 6K+(aq) + 2PO4^3-(aq)
The product, Ca3(PO4)2(s), remains in its solid state, so no dissociation is required.
Lastly, for 6KCl(aq), it dissociates into 6K+ and 6Cl- ions:
6KCl(aq) → 6K+(aq) + 6Cl-(aq)
Now, putting it all together, the total ionic equation becomes:
3Ca2+(aq) + 6Cl-(aq) + 6K+(aq) + 2PO4^3-(aq) → Ca3(PO4)2(s) + 6K+(aq) + 6Cl-(aq)
Note: In the total ionic equation, only the compounds that undergo dissociation are shown as ions, while the solid compound remains intact.
Hope that helps!