Question
Consider the following reaction:
Na2CO3 + NiCl2 ¨ NiCO3 + 2NaCl
Using the solubility rules determine the solubility of all reactants and products. Explain your answer Please.
Which product is the precipitate in this reaction?
Write the complete ionic equation for this reaction:
Write the net ionic equation for this reaction:
Answers
Here is a simplified set of solubility rules. You should memorize this table.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
what is the complete and net ionic equation? I just done understand.... HELP please!
You should have the phase; since you don't I will add what I think is appropriate.
Na2CO3(aq) + NiCl2(aq) ==> NiCO3(s) + 2NaCl(aq)
2Na^+(aq) + CO3^2-(aq) + Ni^2+(aq) + 2Cl^-(aq) ==> NiCO3(s) + 2Na^+(aq) + 2Cl^-(aq)
This is the complete ionic equation. To make it a net ionic equation you go through and cancel ions that are the same on each side of the equation What is left is the net ionic equation. For example, there is a Cl^- on both side; cancel them. Etc.
Na2CO3(aq) + NiCl2(aq) ==> NiCO3(s) + 2NaCl(aq)
2Na^+(aq) + CO3^2-(aq) + Ni^2+(aq) + 2Cl^-(aq) ==> NiCO3(s) + 2Na^+(aq) + 2Cl^-(aq)
This is the complete ionic equation. To make it a net ionic equation you go through and cancel ions that are the same on each side of the equation What is left is the net ionic equation. For example, there is a Cl^- on both side; cancel them. Etc.
So the net equation is CO3^2-(aq)+Ni^2+(aq)----->NiCO3(s)
yes
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