First off, NaCO3 is unbalanced. It should be Na2CO3.
Na2CO3 --> 2 Na+ + CO32-
CuSO4 --> Cu2+ + SO42-
Together only Cu2+ and CO32- react with each other, Sodium and Sulfate don't form a precipitate and are called: tribune ions, ie. don't react and can be left out of the reaction equation (marked by hooked parenthesis [.+.] )
Cu2+ + [SO42-+ 2 Na+] + CO32- --> (CuCO3)precip. solid+ [SO42-+ 2 Na+]
Simpler:
Cu2+aq. + CO32-aq. --> (CuCO3)precip. solid
First Derive molocular equation using oxidation numbers
CuSO4+KOH-->Cu(OH)2+K2SO4
Then Balance CuSO4+2KOH-->Cu(OH)2+K2SO4
Then determine spectator ions by noting which elements do not change oxidation numbers
Remove your spectator ions and you have your net ionic equation
2NaOH+MgSO4 -------> Na2SO4+ Mg(OH)2
There you go ;)
You know, I have the same question and cant find the answer anywhere. Just my luck.
There is no reaction between these chemicals, so there is NO net ionic equation.
2OH^- + Cu^2+ --------> Cu(OH)2!
2I- + Cu+2 = CuI2 (s)
Any net ionic equation.
HP- + H2O(l) ⇌ P2-(aq) + H3O+(aq)
HCHO2 is methanoic acid and is normally written as HCOOH. When reacted with potassium hydroxide (KOH) , it produces potassium methanoate. and water/ HCOOH +KOH = HCOO^-K+ + H2O
HNO3 + KOH ---> KNO3 + H2O
HNO3 + KOH ----> KNO3 + H2O This is the complete equation this is the net ionic equation. Because it is a strong acid (HNO3) and a strong base (KOH) it will always be H+ + OH- -------> H2O
Balanced :2 K + 2 H2O ----> 2 KOH + H2
CuSO4 + 2 KOH = Cu(OH)2 + K2SO4Copper hydroxide is insoluble in water.
HP- + H2O(l) ⇌ P2-(aq) + H3O+(aq)
HCHO2 is methanoic acid and is normally written as HCOOH. When reacted with potassium hydroxide (KOH) , it produces potassium methanoate. and water/ HCOOH +KOH = HCOO^-K+ + H2O
HNO3 + KOH ---> KNO3 + H2O
HNO3 + KOH ----> KNO3 + H2O This is the complete equation this is the net ionic equation. Because it is a strong acid (HNO3) and a strong base (KOH) it will always be H+ + OH- -------> H2O
CuSO4 + 2 KOH = Cu(OH)2(s) + K2SO4Only copper hydroxide is insoluble in water; other compounds are soluble in water.
The chemical equation is:H3PO4 + 3 KOH = K3PO4 + 3 H2O
Balanced :2 K + 2 H2O ----> 2 KOH + H2
Overall ionic equation: K+(aq) +OH-(aq)+ H30+(aq) + Cl-(aq)------>2H20 (l) + K+(aq) + Cl-(aq) Net ionic Equation OH-(aq)+ H30+(aq) ------>2H20 (l)
The product is insoluble green Copper hydroxide which may be separated by simple filtration
I assume KOH is limiting. Balanced equation. KOH + HCl -> KCl + H2O 0.400 moles KOH (1 mole H2O/1 mole KOH)(18.016 grams/1 mole H2O) = 7.21 grams water produced =====================
Plus 1 charge