Balance the equation:
Note: If a chemical species coefficient is "1" then "1" needs to be entered in the field before that species.
__ MnO4– + __H+ + __C2O42- --> __Mn2+ + __H2O + __CO2
2MnO4^-+ ? H^+ + 5C2O4^= ==>2Mn^+2 + 10CO2 + ?H2O
You should be abl to finish.
so 16 H and then 8H20?
right
To balance the given chemical equation:
First, let's count the number of atoms on each side of the equation:
On the left side:
Atoms | Mn | O | H | C
Count | 1 | 4 | 1 | 2
On the right side:
Atoms | Mn | O | H | C
Count | 1 | 6 | 2 | 1
Since we have an unequal number of atoms on both sides, we need to balance them by adding coefficients.
Let's start with Mn. Since there is only one Mn atom on both sides, we can leave it as it is.
Next, let's balance the O atoms. The MnO4– ion has 4 O atoms, so we need to add a coefficient of 4 to H2O on the right side.
__ MnO4– + __H+ + __C2O42- --> __Mn2+ + 4H2O + __CO2
Now, let's balance the C atoms. The C2O42- ion has 2 C atoms, so we need to add a coefficient of 2 to CO2 on the right side.
__ MnO4– + __H+ + __C2O42- --> __Mn2+ + 4H2O + 2CO2
Finally, let's balance the H atoms. The MnO4– ion has 1 H atom, so we need to add a coefficient of 1 to H+ on the left side.
__ MnO4– + 1H+ + __C2O42- --> __Mn2+ + 4H2O + 2CO2
Now, let's balance the charge by adding electrons (e-) as necessary. The MnO4- ion on the left side has a charge of -1, so we need to add one electron (e-) to the left side.
__ MnO4– + 1H+ + __C2O42- + __e- --> __Mn2+ + 4H2O + 2CO2
Finally, let's balance the charges. The Mn2+ ion on the right side has a charge of +2, so we need to add two electrons (2e-) to the right side.
__ MnO4– + 1H+ + __C2O42- + __e- --> __Mn2+ + 4H2O + 2CO2 + 2e-
Now, the equation is balanced:
1 MnO4– + 1H+ + 1C2O42- + 1e- --> 1Mn2+ + 4H2O + 2CO2 + 2e-