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.
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.
yeah sorry it was KMnO4
i just still have no idea what the full equation would look like...
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.
ok thankyou
To write the balanced chemical equations, we first need to write the net ionic equations and then balance them. A net ionic equation shows only the ions that participate in the reaction, eliminating the spectator ions. Let's start with the first reaction:
1. K2MnO4(aq) + H2SO4(aq) + NaHSO3(aq) → ?
First, let's write the balanced molecular equation:
K2MnO4(aq) + H2SO4(aq) + NaHSO3(aq) → ?
Now, we need to write the complete ionic equation, showing all the ions:
2K+(aq) + MnO4-(aq) + 2H+(aq) + SO4^2-(aq) + Na+(aq) + HSO3-(aq) → ?
To write the net ionic equation, we remove the spectator ions (ions which are present on both sides of the equation, not participating in the reaction). In this case, the spectator ions are K+ and SO4^2-:
MnO4-(aq) + 2H+(aq) + Na+(aq) + HSO3-(aq) → ?
Now we can proceed to balance the equation:
2MnO4-(aq) + 5H+(aq) + 6Na+(aq) + 5HSO3-(aq) → 2Mn2+(aq) + 5Na+(aq) + 5SO4^2-(aq) + 3H2O(l)
The balanced chemical equation for the first reaction is:
2K2MnO4(aq) + 5H2SO4(aq) + 6NaHSO3(aq) → 2MnSO4(aq) + 5Na2SO4(aq) + 3H2O(l) + 2K2SO4(aq)
Now, let's move on to the second reaction:
2. K2MnO4(aq) + NaOH(aq) + NaHSO3(aq) → ?
Following the same procedure, we write the balanced molecular equation:
K2MnO4(aq) + NaOH(aq) + NaHSO3(aq) → ?
Next, we write the complete ionic equation:
2K+(aq) + MnO4-(aq) + Na+(aq) + OH-(aq) + Na+(aq) + HSO3-(aq) → ?
Then, we remove the spectator ions (K+ and Na+):
MnO4-(aq) + OH-(aq) + HSO3-(aq) → ?
Now let's balance the equation:
2MnO4-(aq) + 3OH-(aq) + 3HSO3-(aq) → 2MnO2(s) + 3SO4^2-(aq) + 3H2O(l)
The balanced chemical equation for the second reaction is:
2K2MnO4(aq) + 3NaOH(aq) + 3NaHSO3(aq) → 2MnO2(s) + 3Na2SO4(aq) + 3H2O(l)