This is an oxidation-reduction (redox) reaction:

* 2 MnVII + 10 e- → 2 MnII (reduction)*

* 15 C-II - 10 e- → 15 C-4/3 (oxidation)*

** KMnO4** is an oxidizing agent,

**is a reducing agent.**

*C3H8O*### input interpretation

*CH3CH2CH2OH + KMnO4 + H2SO4 → CH3COCH3 + MnSO4 + H2O + K2SO**N-propanol potassium sulfuric acid acetone manganese(II)sulfate potassium *

* Permanganate sulfate *

### Word equation

*N-propanol + potassium permanganate + sulfuric acid acetone + manganese(II) sulfate + water + potassium sulfate*

### Balanced equation

Balance the chemical equation algebraically:

*CH3CH2CH2OH + KMnO4 + H2SO4 → CH3COCH3 + MnSO4 + H2O + K2SO*

Add stoichiometric coefficients, c, to the reactants and products:

*c1**CH3CH2CH2OH + c2KMnO4 + c3H2SO4 → c4 CH3OCH3 + c5MnSO4 + *

* c6H2O + c7K2SO*

Set the number of atoms in the reactants equal to the number of atoms in the products for ** C, H, O, K, Mn **and

*S:**C: 3c1 = 3c4*

*H: 8c1+ 2c3 = 6c4 +2c6*

*O: c1+ 4c2 + 4c3 = c4 + 4c5 + c6 + 4c7*

*K: c2 = 2c7*

*Mn: c2 = c5*

*S: c3 = c5 + c7*

Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set ** c7 = 1 **and solve the system of equations for the remaining coefficients:

* c1 = 5 *

* c2=2*

*c3 = 3 *

*c4=5*

*c5=2 *

*c6=8*

*c7=1*

Substitute the coefficients into the chemical reaction to obtain the balanced equation:

**Answer**:

*CH3CH2CH2OH + KMnO4 + H2SO4 → CH3COCH3 + MnSO4 + H2O + K2SO*