Please help balance this:
CH3COO + CrO4 + H = Cr(OH)3 + CO2 + H2O
To balance the chemical equation:
CH3COO + CrO4 + H → Cr(OH)3 + CO2 + H2O
1. Start by counting the number of atoms on each side of the equation.
On the left side:
- Carbon (C): 2
- Hydrogen (H): 4
- Oxygen (O): 5
On the right side:
- Chromium (Cr): 1
- Oxygen (O): 10
- Hydrogen (H): 7
- Carbon (C): 1
2. Balance the atoms that appear in the fewest number. In this case, it's Chromium (Cr) on the right side. Balance it by putting a coefficient of 8 in front of Cr(OH)3:
CH3COO + CrO4 + H → 8Cr(OH)3 + CO2 + H2O
3. Now check the number of Hydrogen (H) atoms.
On the left side: 4
On the right side: 56 (8 multiplied by 7)
To balance the Hydrogen, put a coefficient of 14 in front of H on the left side:
CH3COO + 14H + CrO4 → 8Cr(OH)3 + CO2 + H2O
4. Next, balance the Oxygen (O) atoms:
On the left side: 5
On the right side: 88 (8 multiplied by 10, plus 2)
To balance the Oxygen, put a coefficient of 23/2 (or 11.5) in front of CrO4 on the left side:
CH3COO + 14H + (11.5)CrO4 → 8Cr(OH)3 + CO2 + H2O
Note: In this case, you can use fractional coefficients because the reaction is taking place in a solution and not in strict stoichiometric ratios.
5. Finally, balance the Carbon (C) atoms:
On the left side: 1
On the right side: 8
To balance the Carbon, put a coefficient of 8 in front of CH3COO on the left side:
8CH3COO + 14H + (11.5)CrO4 → 8Cr(OH)3 + CO2 + H2O
Now, the equation is balanced with the same number of atoms on both sides.