Hi, I balanced these already but I'm having trouble writing the total and net ionic reaction.
1) 3Mg + 2Al(NO3)3(aq) -> 2Al + 3Mg(NO3)2
2) Cu + 2AgNO3(aq) -> 2Ag + Cu(NO3)2
3) Fe + 2Cu(NO3)2(aq) -> Fe(NO3)2+2Cu
4) 2Fe + 3H2SO4(aq) -> Fe2(SO4)3+3H2
Writing total ionic reaction and net ionic reaction?
2017-03-24 5:21 pm
回答 (2)
2017-03-24 9:47 pm
✔ 最佳答案
1.Total ionic equation: 3Mg(s) + 2Al²⁺(aq) + 6NO₃⁻(aq) → 2Al(s) + 3Mg²⁺(aq) + 6NO₃⁻(aq)
Net ionic equation: 3Mg(s) + 2Al²⁺(aq) → 2Al(s) + 3Mg²⁺(aq)
2.
Total ionic equation: Cu(s) + 2Ag⁺(aq) + 2NO₃⁻(aq) → 2Ag(s) + Cu²⁺(aq) + 2NO₃⁻(aq)
Net ionic equation: Cu(s) + 2Ag⁺(aq) → 2Ag(s) + Cu²⁺(aq)
3.
The given equation should be Fe + Cu(NO₃)₂(aq) → Fe(NO₃)₂ + Cu instead.
Total ionic equation: Fe(s) + Cu²⁺(aq) + 2NO₃⁻(aq) → Fe²⁺(aq) + 2NO₃⁻(aq) + Cu(s)
Net ionic equation: Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s)
4.
Total ionic equation: 2Fe(s) + 6H⁺(aq) + 3SO₄²⁻(aq) → 2Fe²⁺(aq) + 3SO₄²⁻(aq) + 3H₂(aq)
Net ionic equation: 2Fe(s) + 6H⁺(aq) → 2Fe²⁺(aq) +3H₂(aq)
2017-03-24 8:50 pm
1) 3Mg(s) + 2Al^3+(aq) + 6NO3^2-(aq) ==> 2Al(s) + 3Mg^2+(aq) + 6NO3^2-(aq) total ionic
3Mg(s) + 2Al^3+(aq) ==> 2Al(s) + 3Mg^2+(aq) net ionic (after removing spectator ions)
2) same thing just substitute Cu and Ag for Mg and Al (NO3 is a spectator)
You should be able to do the others. If you have a precipitate, a non electrolyte/weak electrolyte or a gas, they should appear in the net ionic
3Mg(s) + 2Al^3+(aq) ==> 2Al(s) + 3Mg^2+(aq) net ionic (after removing spectator ions)
2) same thing just substitute Cu and Ag for Mg and Al (NO3 is a spectator)
You should be able to do the others. If you have a precipitate, a non electrolyte/weak electrolyte or a gas, they should appear in the net ionic
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