The following reactions result in formation of a complex ion. Write a balanced m

Question

The following reactions result in formation of a complex ion. Write a balanced molecular equation and then the net-ionic equation for each. Assume reactants are aqueous solutions unless underlined (in which case they are solids). Show the physical form of all species (e.g., (aq), (s), etc.). (xs = excess reactant) zinc chloride + sodium hydroxide (xs) rightarrow silver(I) nitrate + potassium cyanide (xs) rightarrow lead (II) hydroxide + sodium hydroxide (xs) rightarrow tin(II) nitrate + hydrochloric acid (xs) rightarrow nickel (II) hydroxide + ammonia (xs) rightarrow

Explanation / Answer

1. Zinc Chloride and Sodium hydroxide:

Reaction: ZnCl2 (aq) + 2NaOH (aq) = 2NaCl (aq) + Zn(OH)2 (S)

Rewrite the equation with showing all the ions:

Zn 2+(aq) + 2Cl-(aq) + Na 2+(aq) + 2OH-(aq) = Na 2+(aq) + 2Cl-(aq) + Zn(OH)2 (s)

Since, sodium and chlorine ions are present on both sides of the reaction and are not changed by the reaction, so you can cancel them from both sides of the reaction. This leaves you with the net ionic equation:

Zn(2+)(aq) + 2OH(-) (aq) = Zn(OH)2(s)

2. Silver nitrate and potassium cyanide:

AgNO3(aq) + 2 KCN(aq) ---> K[Ag(CN)2](aq) + KNO3(aq)

3. Lead hydroxide and sodium hydroxide:

Since, both lead hydroxide [Pb(OH)2] and sodium hydroxide (NaOH) have the same anion, OH-There won't be any reaction in this case.

4. tin(ii)nitrate and hydrochoric acid:

Reaction: Sn(NO3)2 + 2HCL = SnCl2 (s) + 2HNO3

We can rewrite the net ionic equation as we have done in reaction 1

net ionic equation:

Sn 2+ (aq) + 2 Cl- (aq) = PbCl2(s)

5. Nicket(ii) hydroxide and ammonia:

I am not Confirmed for this reaction.

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