Plz help me balance this chemical eqn H2C2O4.2H2O+KMnO4
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I assume this is an acidic solution. And what kind of acid do you want to use? I'll show HCl.
5H2C2O4.2H2O + 2KMnO4 + 6HCl ==> 2MnCl2 + 18H2O + 10CO2 + 2KCl
To balance the chemical equation H2C2O4.2H2O + KMnO4, you need to ensure that the same number of atoms of each element are present on both sides of the equation. To do this, follow these steps:
Step 1: Count the number of atoms of each element on both sides of the equation.
H2C2O4.2H2O + KMnO4
On the left side:
- Hydrogen (H): 4 atoms from H2C2O4 and 4 atoms from 2H2O, so a total of 8 atoms.
- Carbon (C): 2 atoms from H2C2O4.
- Oxygen (O): 12 atoms from H2C2O4 and 4 atoms from 2H2O, so a total of 16 atoms.
On the right side:
- Hydrogen (H): 0 atoms.
- Carbon (C): 0 atoms.
- Oxygen (O): 4 atoms from KMnO4.
Step 2: Balance the least abundant element, which in this case is carbon (C).
To balance the carbon, place a coefficient of 2 in front of H2C2O4 to have 2 carbon atoms on each side:
2 H2C2O4.2H2O + KMnO4
Now, we have:
- Hydrogen (H): 8 atoms.
- Carbon (C): 4 atoms.
- Oxygen (O): 16 atoms.
Step 3: Balance the hydrogen (H) atoms.
The hydrogen atoms are already balanced on both sides.
Now, we have:
- Hydrogen (H): 8 atoms.
- Carbon (C): 4 atoms.
- Oxygen (O): 16 atoms.
Step 4: Balance the oxygen (O) atoms.
To balance the oxygen atoms, add a coefficient of 7/2 in front of KMnO4 to have 16 oxygen atoms on each side. Since coefficients should be whole numbers, multiply all coefficients by 2 to eliminate the fraction:
4 H2C2O4.2H2O + 7 KMnO4
Now, we have:
- Hydrogen (H): 8 atoms.
- Carbon (C): 4 atoms.
- Oxygen (O): 16 atoms.
Step 5: Verify the balance of all atoms.
- Hydrogen (H): 8 atoms on both sides.
- Carbon (C): 4 atoms on both sides.
- Oxygen (O): 16 atoms on both sides.
Therefore, the balanced chemical equation is:
4 H2C2O4.2H2O + 7 KMnO4 -> 8 H2O + 4 CO2 + 7 MnO2 + K2O