write the net ionic equation for the following :
2CuCl(aq)-> CuCl2(aq)+ Cu(s)
4NaCl(aq)+2H2SO4(aq)+MnO2(s)-> 2Na2SO4(aq)+2H2O(l)+Cl2(g)
3KOH(aq)+Fe(NO3)3(aq)->Fe(OH)3(s)+ 3KNO3(aq)
I don't believe #1 since CuCl is not very soluble in water; however, since it "says" aq, we will do it as is.
2CuCl(aq) --> CuCl2(aq) + Cu(s)
2Cu^+(aq) + 2Cl^-(aq) ==>Cu^+2(aq) + 2Cl^-(aq) + Cu(s)
Now cancel ions common to both sides (Cl^-) to arrive at the net equation.
2Cu^+(aq) ==>Cu^+2(aq) + Cu(s)
Do the others the same way.
1. Balance the molecular equation.
2. Convert to the ionic equation. Solids are written as the molecule, weak electrolytes (such as water, etc) are written as the molecule, gases are written as the molecule. All others are written as ions.
3. Cancel ions common to both sides to arrive at the net ionic equation.
To write the net ionic equation, we need to first write the balanced molecular equation and then identify the spectator ions, which do not participate in the reaction.
1) 2CuCl(aq) -> CuCl2(aq) + Cu(s)
The balanced molecular equation is:
2CuCl(aq) -> CuCl2(aq) + Cu(s)
To write the net ionic equation, we omit the spectator ions. In this case, the spectator ion is Cl-.
Net ionic equation:
2Cu+(aq) + 2e- -> Cu(s)
2) 4NaCl(aq) + 2H2SO4(aq) + MnO2(s) -> 2Na2SO4(aq) + 2H2O(l) + Cl2(g)
The balanced molecular equation is:
4NaCl(aq) + 2H2SO4(aq) + MnO2(s) -> 2Na2SO4(aq) + 2H2O(l) + Cl2(g)
To write the net ionic equation, we omit the spectator ions. In this case, the spectator ions are Na+ and SO4^2-.
Net ionic equation:
2H+(aq) + MnO2(s) -> H2O(l) + Cl2(g)
3) 3KOH(aq) + Fe(NO3)3(aq) -> Fe(OH)3(s) + 3KNO3(aq)
The balanced molecular equation is:
3KOH(aq) + Fe(NO3)3(aq) -> Fe(OH)3(s) + 3KNO3(aq)
To write the net ionic equation, we omit the spectator ions. In this case, the spectator ions are K+ and NO3-.
Net ionic equation:
3OH-(aq) + Fe^3+(aq) -> Fe(OH)3(s)
To write the net ionic equation for a chemical reaction, you need to first write the balanced molecular equation and then the ionic equation. Finally, you can simplify the ionic equation to obtain the net ionic equation.
1. Molecular Equation:
2CuCl(aq) → CuCl2(aq) + Cu(s)
2. Ionic Equation:
To write the ionic equation, you need to break down the reactants and products into their respective ions:
2Cu2+(aq) + 4Cl-(aq) → Cu2+(aq) + 2Cl-(aq) + Cu(s)
3. Net Ionic Equation:
In the net ionic equation, you eliminate the spectator ions (ions that appear on both sides of the equation, unchanged). In this case, the Cu2+ ion is present on both sides, so it is a spectator ion.
4Cl-(aq) → 2Cl-(aq) + Cu(s)
Therefore, the net ionic equation for the reaction 2CuCl(aq) → CuCl2(aq) + Cu(s) is 4Cl-(aq) → 2Cl-(aq) + Cu(s).
Now, let's move on to the other examples:
1. Molecular Equation:
4NaCl(aq) + 2H2SO4(aq) + MnO2(s) → 2Na2SO4(aq) + 2H2O(l) + Cl2(g)
2. Ionic Equation:
4Na+(aq) + 4Cl-(aq) + 2H+(aq) + SO4^2-(aq) + MnO2(s) → 2Na+(aq) + 2SO4^2-(aq) + 2H2O(l) + Cl2(g)
3. Net Ionic Equation:
4Cl-(aq) + 2H+(aq) + MnO2(s) → 2H2O(l) + Cl2(g)
The net ionic equation for the given chemical reaction is 4Cl-(aq) + 2H+(aq) + MnO2(s) → 2H2O(l) + Cl2(g).
Moving on to the last example:
1. Molecular Equation:
3KOH(aq) + Fe(NO3)3(aq) → Fe(OH)3(s) + 3KNO3(aq)
2. Ionic Equation:
3K+(aq) + 3OH-(aq) + Fe^3+(aq) + 3NO3-(aq) → Fe(OH)3(s) + 3K+(aq) + 3NO3-(aq)
3. Net Ionic Equation:
3OH-(aq) + Fe^3+(aq) → Fe(OH)3(s)
The net ionic equation for the given chemical reaction is 3OH-(aq) + Fe^3+(aq) → Fe(OH)3(s).