Write the net ionic equation (including phases) that corresponds to pb (no3)2(aq)+k2s(aq)→pbs(s)+2kno3(aq)

Write the net ionic equation (including phases) that corresponds to pb (no3)2(aq)+k2s(aq)→pbs(s)+2kno3(aq)

Answers

                       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)

Pb^{2+}(aq)+2S^{2-}(aq)rightarrow PbS(s)

Explanation:

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:

Pb(NO_3)_2(aq)+K_2S(aq)rightarrow PbS(s)+2KNO_3(aq)

Pb^{2+}(aq)+2NO_3^-(aq)+2K^+(aq)+S^{2-}(aq)rightarrow PbS(s)+2K^+(aq)+NO_3^-(aq)

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 Pb^{2+}(aq)+2S^{2-}(aq)rightarrow PbS(s)

1) The question is incomplete. What is the compound which Fe(ClO4)2 reacts with?

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.

Explanation:

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:

Zn(NO_3)_2(aq.)+Na_2CO_3(aq.)rightarrow ZnCO_3(s)+2NaNO_3(aq.)

Ionic form of the above equation follows:

Zn^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+CO_3^{2-}(aq.)rightarrow ZnCO_3(s)+2Na^+(aq.)+2NO_3^-(aq.)

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:

Zn^{2+}(aq.)+CO_3^{2-}(aq.)rightarrow ZnCO_3(s)

Hence, the net ionic equation is written above.

Complete ionic:
Cu(aq) + 2Cl(aq) + 8O(aq) + 2Na(aq) + C(aq) + 3O(aq) = CaCO3(s) + 2Na(aq) + Cl(aq) + 4O(aq) 

Net ionic:

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! 

KOH(aq) → KCN(aq) + H2O(l)
HCN(aq) + OH-(aq) → CN-(aq) + H2O(l)
The net ionic equation that correspond to Fe(ClO4)2 +Na2CO3  is

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)

The net ionic equation that corresponds to Zn(ClO4)2(aq)+K2S(aq)= ZnS(S)+2KClO4(aq) the answer is Zn2+(aq) + 2CO3^2-(aq) --> Zn(CO3)2(s). Thank you for posting your question here. I hope my answer helps.
Zn(no3)2  +  NaCO3  >  ZnCO3  +   2NaNO3

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-
this  yield
Zn^2+(aq)   +  CO^-2(aq)  ZnCO3(s)

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