How would you balance this equation to find the coefficients of each compound and the spectator ion ?
KI(aq)+ KMnO4 (aq) + H2O= I2(s) + MnO2(s) + KOH(aq)
6KI + 2KMnO4 + 4H2O → 3I2 + 2MnO2 + 8KOH
6KI + 2KMnO4 + 4H2O → 3I2 + 2MnO2 + 8KOH
To find the spectator ions, first divide the individual reactants/products into their respetive ions.
6K^+ + 6I^- + 2K^+ + 2MnO4^- + 4H2O → 3I2 + 2MnO2 + 8K^+ + 8OH^-
Then note the ions present at the end. The ions that apear on both sides of the equation are the spectator ions. K^+ ions are the spectator ions.
To balance this chemical equation, we need to make sure that there is an equal number of atoms on both sides. Here's how you can balance this equation step by step:
Step 1: Count the number of each type of atom on both sides of the equation.
On the reactant side:
- K: 1
- I: 1
- Mn: 1
- O: 4
- H: 2
On the product side:
- I: 2
- Mn: 1
- O: 2
- H: 1
- K: 1
Step 2: Identify the elements that are not balanced. In this case, we have I (iodine) and O (oxygen).
Step 3: Balance the least abundant atom first. In this case, Mn (manganese) is the least abundant. Add a coefficient of 6 in front of KMnO4 to balance the Mn atoms.
6 KI(aq) + 6 KMnO4 (aq) + H2O → I2(s) + 6 MnO2(s) + KOH(aq)
Step 4: Now, let's balance the iodine atoms (I). Since there are 6 I atoms on the reactant side and only 2 on the product side, we can add a coefficient of 3 in front of I2 to balance the I atoms.
6 KI(aq) + 6 KMnO4 (aq) + H2O → 3 I2(s) + 6 MnO2(s) + KOH(aq)
Step 5: Balance the oxygen atoms (O). Starting with the products, there are 6 O atoms in MnO2 and 1 O atom in KOH, giving us a total of 7 O atoms. On the reactant side, there are 7 O atoms in the 6 KMnO4 molecules. Therefore, we don't need to change anything since O is already balanced.
Final Balanced Equation:
6 KI(aq) + 6 KMnO4 (aq) + H2O → 3 I2(s) + 6 MnO2(s) + KOH(aq)
The coefficients of each compound are: 6, 6, 1, 3, 6, 1.
The spectator ion is: K+ (potassium ion).