Molecular,ionic,net ionic equations-
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)
