Complete and balance the following equation. Then write it as a net ionic equation. K3PO4(aq) + MgCl2(aq)
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2 K3PO4 (aq) + 3 MgCl2 (aq) → Mg3(PO4)2 (s) & 6 KCl (aq)
Net Ionic
2(PO4)-3 (aq) + 3 Mg+2 (aq) → Mg3(PO4)2 (s)
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what is the complete ionic equation for 2K3PO4(aq)+3MgCl2(aq) → Mg3(PO4)2(s)+6KCl(aq)
6K+(aq)+2PO43−(aq)+3Mg2+(aq)+6Cl−(aq)→Mg3(PO4)2(s)+6K+(aq)+6Cl−(aq)
complete ionic equation
To complete and balance the equation, we need to find the products formed when K3PO4(aq) and MgCl2(aq) react. First, let's write the equation:
K3PO4(aq) + MgCl2(aq) -> ?
We can determine the products by swapping the cations (positively charged ions) and anions (negatively charged ions) in the reactants. In this case, the cation K+ from K3PO4 will combine with the anion Cl- from MgCl2, and vice versa.
K3PO4(aq) + MgCl2(aq) -> KCl + Mg3(PO4)2
Now, let's balance the equation by ensuring that the number of atoms on both sides of the equation is equal. We can start by balancing the phosphorus (P) and chloride (Cl) atoms:
K3PO4(aq) + 3MgCl2(aq) -> 3KCl + Mg3(PO4)2
The equation is now balanced.
To write this equation as a net ionic equation, we need to represent all the species that dissociate (ionize) in the solution into their individual ions. In this case, both K3PO4 and MgCl2 dissociate completely, whereas the products KCl and Mg3(PO4)2 do not dissociate significantly in aqueous solution.
The net ionic equation would be:
3K+ (aq) + PO4^3- (aq) + 6Cl- (aq) + 6H2O (l) -> 3KCl (aq) + Mg3(PO4)2 (s)
In the net ionic equation, we only include the ions that participate in the reaction, and any non-ionized substances are represented as they are.
Hope this helps!