Asked by izzy
Write the balanced chemical equation for each reaction:
K2MnO4(aq)+H2SO4(aq)+NaHSO3(aq)= ?
K2MnO4(aq)+NaOH(aq)+NaHSO3(aq)= ?
start with the net ionic equations.
K2MnO4(aq)+H2SO4(aq)+NaHSO3(aq)= ?
K2MnO4(aq)+NaOH(aq)+NaHSO3(aq)= ?
start with the net ionic equations.
Answers
Answered by
DrBob222
Do you mean K2MnO4 or KMnO4? If K2MnO4 then Mn is +6. If KMnO4 then Mn is +7.
In acid solution, Mn goes to +2 (Mn^+2), in basic solution to MnO2. S in HSO3^- goes to +6 (SO4^-2) in acid or basic solution.
In acid solution, Mn goes to +2 (Mn^+2), in basic solution to MnO2. S in HSO3^- goes to +6 (SO4^-2) in acid or basic solution.
Answered by
izzy
yeah sorry it was KMnO4
Answered by
izzy
i just still have no idea what the full equation would look like...
Answered by
DrBob222
You do the half equations.
MnO4^- ==> Mn^+2
HSO3^- ==> SO4^-2
For example, the permanganate half equation is
5e +MnO4^- + 8H^+ ==> Mn^+2 + 4H2O
Do the HSO3^-, multiply by a number to make the electrons equal, then add the two equations. This gives you the ionic equation for that reaction. Then add K^+ and Na^+ as needed for the positive charges and add SO4^-2 (from H2SO4) to balance those charges.
MnO4^- ==> Mn^+2
HSO3^- ==> SO4^-2
For example, the permanganate half equation is
5e +MnO4^- + 8H^+ ==> Mn^+2 + 4H2O
Do the HSO3^-, multiply by a number to make the electrons equal, then add the two equations. This gives you the ionic equation for that reaction. Then add K^+ and Na^+ as needed for the positive charges and add SO4^-2 (from H2SO4) to balance those charges.
Answered by
izzy
ok thankyou
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