Write the net ionic equation (including phases) that corresponds to pb (no3)2(aq)+k2s(aq)→pbs(s)+2kno3(aq)
Pb⁺(aq) + 2NO₃⁻(aq) + K₂⁺(aq) + S⁻(aq) → PbS(s) + K₂⁺(aq) + 2NO₃⁻(aq)
Pb⁺(aq) + S⁻(aq) → PbS(s)
Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.
The given chemical equation is:
The ions which are present on both the sides of the equation are potassium and nitrate ions and hence are not involved in net ionic equation is:
Hence, the net ionic equation is
It could be one of several salts.
2) Assuming it is Na2S the complete reaction is:
Fe(ClO4)2 (aq) + Na2S(aq) → FeS (s) + 2NaClO4 (aq)
So, the next steps how how to work this problem assuming that reaction.
3) Show the ionic compounds as separate ions.
Fe(2+) (aq) + 2 ClO4(-) (aq) + 2 Na(+) (aq) + S(2-) (aq) → FeS(s) + 2 Na(aq) + 2 ClO4(-) (aq)
That is the total ionic equation.
4) Cancel the ions that appear on both sides of the equation , Na(+) and ClO4(-).
Fe(2+) (aq) + S(2-) (aq) → FeS(s) < this is the net ionic equation
What happens to sodium and perchlorate is that they do not participate in the reaction but remain dissolved which is called "spectator ions".
The net ionic equation is written below.
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of zinc nitrate and sodium carbonate is given as:
Ionic form of the above equation follows:
As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
Hence, the net ionic equation is written above.
Cu(aq) + 2Cl(aq) + 8O(aq) + 2Na(aq) + C(aq) + 3O(aq) = CaCO3(s) + 2Na(aq) + Cl(aq) + 4O(aq)
Cu(aq) + Cl(aq) + 4O(aq) + 2Na(aq) + C(aq) + 3O(aq) = CaCO3(s)
So write everything out as IF it will dissociate in water. So everything that is aq splits but solid just floats to the bottom of the mixture. Cancel what you can (in this case the two from the ClO4 on the left of the equation cancels with the ClO4 from the right) and the 2Na cancels. Then, write out the whole solution and you are done!
HCN(aq) + OH-(aq) → CN-(aq) + H2O(l)
Fe ^2+(aq) +CO3^2-(aq) → FeCO3 (s)
Explanationwrite the balance chemical equation
Fe(ClO4)2 (aq) + Na2CO3(aq)→ FeCO3 (s) + 2 NaClO4
write the ionic equation
Fe^2+(aq) + 2ClO4 ^-(aq) + 2Na^+(aq) +CO3^2- (aq) → FeCO3 (s) + 2Na^+ + 2ClO4^-
cancel the spectator ions in both sidethat is ( 2ClO4^- and 2Na+)
the ionic equation is therefore
Fe^2+(aq) + CO3^2- →FeCo3(s)
Zn^2+(aq) + 2 NO3^-(aq) + 2Na^+(aq) + CO3^2(aq)- ---> ZnCO3(s) + 2Na^+(aq) + 2NO3^-(aq)
we cancel the the spectator in both side in this case is Na^+ and NO3-
Zn^2+(aq) + CO^-2(aq) ZnCO3(s)