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When aqueous solutions of Pb(NO3)2(aq) and KI(aq) are mixed, the products are KNO3(aq) and PbI2(s). What is the net ionic equation for this reaction?Pb(NO3)2(aq)+2KI(aq)⟶PbI2(s)+2KNO3(aq)Pb2+(aq)+2NO−3(aq)+2K+(aq)+2I−(aq)⟶PbI2(s)+2K+(aq)+2NO−3(aq)Pb2+(aq)+2I−(aq)⟶PbI2(s)2K+(aq)+2NO−3(aq)⟶2KNO3(s)

Sagot :

The net ionic equation for the reaction that occurs when solutions of lead(II) nitrate[Pb(NO3)2 ] and potassium iodide(2KI) are combined is:

Pb²⁺(aq) + 2 I⁻(aq) ⇒ PbI₂(s)

The balanced Molecular equation for the reaction of lead nitrate [Pb(NO3)2 ]and Potassium iodide(2KI) is given as ;

Pb(NO3)2 (l) + 2KI (s) → PbI2 (s) + 2KNO3 (l).

Solutions of lead(II) nitrate and potassium iodide were combined and formed double displacement reaction. The complete ionic equation is:

Pb²⁺(aq) + 2 NO₃⁻(aq) + 2 K⁺(aq) + 2 I⁻(aq) ⇒ 2 K⁺(aq) + 2 NO₃⁻(aq) + PbI₂(s)

Removing Spectator ions, that appear both in the reactants and in the products. They do not participate in the reaction thats why they do not appear in the net ionic equation.So, The net ionic equation is:

Pb²⁺(aq) + 2 I⁻(aq) ⇒ PbI₂(s)

The net ionic equation for the reaction of lead(II) nitrate[Pb(NO3)2 ] and potassium iodide(2KI) solution is combined:

Pb²⁺(aq) + 2 I⁻(aq) ⇒ PbI₂(s)

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