Molecular,ionic,net ionic equations-

net ionic

Source: quizlet.com

 Balance the given molecular equation & give total ionic and net ionic equations & also give their physical state-

Problem 1-  NaCl(aq) + AgNO3(aq) —> NaNO3(aq) +AgCl(s)

Balanced equation is,

NaCl(aq) + AgNO3(aq) —> NaNO3(aq) +AgCl(s)

Total ionic equation,

Na(aq) + Cl(aq) + Ag+(aq) + NO3(aq) —> Na+ (aq) + NO3(aq) +AgCl(s)

Net ionic equation,

Ag+(aq)+ Cl(aq) —-> AgCl(s)

 

Problem 2 –

BaCl2(aq) + AgNO3(aq) —> Ba(NO3)2(aq) +AgCl(s)

Balanced equation is,

BaCl2(aq)  + 2AgNO3(aq) —> Ba(NO3)2(aq) + 2AgCl(s)

Total ionic equation,

Ba+2 (aq) + 2Cl(aq) + 2Ag+(aq) + 2NO3(aq) —> Ba+2 (aq) + 2 NO3(aq) + 2AgCl(s)

Ba+2  & NO3–  are spectator ions ,so cancelled out.

Net ionic equation,

Ag+(aq)+ Cl(aq) —-> AgCl(s)

 

Problem 3 –

CuCl2(aq) + K3PO4(aq) —>  Cu3(PO4)3(s)  + KCl (aq)

Balanced equation is,

3CuCl2(aq) + 2K3PO4(aq) —>  Cu3(PO4)2(s)  + 6KCl (aq)

Total ionic equation,

3Cu+2 (aq) + 6Cl(aq) + 6K+(aq) + 2PO4-3 (aq) —> Cu3(PO4)2(s)  + 6K+(aq)  + 6Cl(aq)

K+ &  Cl are spectator ions ,so cancelled out.

Net ionic equation,

3Cu+2 (aq) + 2PO4-3 (aq) —> Cu3(PO4)2(s)

 

Problem 4 –

Pb(NO3)2 (aq) + KCl (aq) —> PbCl2(s)  + KNO3(aq)

Balanced equation is,

Pb(NO3)2 (aq) + 2KCl (aq) —> PbCl2(s)  + 2KNO3(aq)

Total ionic equation,

Pb+2 (aq) + 2NO3(aq) + 2K+(aq) + 2Cl(aq) —> PbCl2(s)  + 2K+(aq)  +2NO3(aq)

K+ &  NO3 are spectator ions ,so cancelled out.

Net ionic equation,

Pb+2 (aq) + 2Cl(aq) —> PbCl2(s)

 

Problem 5 –

Pb(NO3)2 (aq) + KI (aq) —> PbI2(s)  + KNO3(aq)

Balanced equation is,

Pb(NO3)2 (aq) + 2KI (aq) —> PbI2(s)  + 2KNO3(aq)

Total ionic equation,

Pb+2 (aq) + 2NO3(aq) + 2K+(aq) + 2I(aq) —> PbI2(s)  + 2K+(aq)  +2NO3(aq)

K+ &  NO3 are spectator ions ,so cancelled out.

Net ionic equation,

Pb+2 (aq) + 2I(aq) —> PbI2(s)

Problem 6-

Pb(ClO4)2 (aq) + KCl (aq) —> PbCl2(s)  + KClO4 (aq)

Balanced equation is,

Pb(ClO4)2 (aq) + 2KCl (aq) —> PbCl2(s)  + 2KClO4 (aq)

Total ionic equation,

Pb+2 (aq) + 2ClO4(aq) + 2K+(aq) + 2Cl(aq) —> PbCl2(s)  + 2K+(aq)  +2ClO4(aq)

K+ &  ClO4  are spectator ions ,so cancelled out.

Net ionic equation,

Pb+2 (aq) +2Cl(aq) —> PbCl2(s)

Problem 7 –

Bi(NO3)2 (aq) + BaS (aq) —> Bi2S3(s)  + Ba(NO3)2(aq)

Balanced equation is,

2Bi(NO3)3 (aq) + 3BaS (aq) —> Bi2S3(s)  + 3Ba(NO3)2(aq)

Total ionic equation,

2Bi+3 (aq) + 6NO3(aq) + 3Ba+2(aq) + 3S-2 (aq) —> Bi2S3(s)  + 3Ba+2 (aq)  +6NO3(aq)

Ba+2 &  NO3 are spectator ions ,so cancelled out.

Net ionic equation,

2Bi+3 (aq) + 3S-2 (aq) —> Bi2S3(s)