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(+) + __C2O4(2–) __Mn(2+) + __H2O + __CO2
i need a check pls
on this one i got:2,16,2,2,8,4 but it say that the second(2)which is the coefficent for C204(2-) and the 4 on the co2 is wrong ? what might be the problem?
need the check pls
Mn goes from +7 on the left to +2 on the right. C goes from +6 on the left (for both C atoms) to +8 on the right (for 2 C atoms). [Place a 2 in front of CO2 to start]. To make the electron loss and gain equal, multiply the Mn half reaction by 2 and the C2O4/2CO2 half reaction by 5 (10 e gained=10 e lost)
Of course the 2 and 5 coefficients balance only the redox part of the equation. You must still balance the H^+ and H2O by inspection.
To balance the given chemical equation, follow these steps:
1. Start by balancing the atoms that appear in the equation except for hydrogen and oxygen. In this case, we have:
__MnO4(-) + __H(+) + __C2O4(2-) → __Mn(2+) + __H2O + __CO2
Manganese (Mn) is balanced with one Mn atom on both sides.
2. Balance the carbon atoms next. We have:
__C2O4(2-) → __CO2
To balance the carbon atoms, we need to have two CO2 molecules on the right side:
__C2O4(2-) → 2CO2
3. Now, balance the oxygen atoms. Start from the left side:
MnO4(-) → Mn(2+)
Since there are four oxygen atoms on the left side and one on the right side, we need to add three water molecules (H2O) to the right side to balance the oxygen atoms:
MnO4(-) → Mn(2+) + 3H2O
4. Next, balance the hydrogen atoms. We have:
H(+) + H2O → __H2O + __H(+)
To balance the hydrogen atoms, we add eight H(+) ions to the left side:
8H(+) + MnO4(-) → Mn(2+) + 3H2O
5. Finally, check if all the atoms on both sides are balanced. We have:
On the left side: 1 Mn, 8 H, 4 O, 2 C
On the right side: 1 Mn, 4 H, 8 O, 2 C
All the atoms are now balanced.
Therefore, the balanced equation is:
8H(+) + MnO4(-) + C2O4(2-) → Mn(2+) + 3H2O + 2CO2