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, C3H8O is a reducing agent.
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:
c1CH3CH2CH2OH + 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
