KOH + MnO2 + KClO3 = H2O + KMnO4 + KCl

potassium hydroxide + manganese dioxide + potassium chlorate ⟶ water + potassium permanganate + potassium chloride

Balance the chemical equation algebraically: + + ⟶ + + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 + c_3 ⟶ c_4 + c_5 + c_6 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Mn and Cl: H: | c_1 = 2 c_4 K: | c_1 + c_3 = c_5 + c_6 O: | c_1 + 2 c_2 + 3 c_3 = c_4 + 4 c_5 Mn: | c_2 = c_5 Cl: | c_3 = c_6 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 c_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 2 c_3 = 1 c_4 = 1 c_5 = 2 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 + 2 + ⟶ + 2 +

+ + ⟶ + +

potassium hydroxide + manganese dioxide + potassium chlorate ⟶ water + potassium permanganate + potassium chloride

| potassium hydroxide | manganese dioxide | potassium chlorate | water | potassium permanganate | potassium chloride molecular free energy | -379.4 kJ/mol | -465.1 kJ/mol | -296.3 kJ/mol | -237.1 kJ/mol | -737.6 kJ/mol | -408.5 kJ/mol total free energy | -758.8 kJ/mol | -930.2 kJ/mol | -296.3 kJ/mol | -237.1 kJ/mol | -1475 kJ/mol | -408.5 kJ/mol | G_initial = -1985 kJ/mol | | | G_final = -2121 kJ/mol | | ΔG_rxn^0 | -2121 kJ/mol - -1985 kJ/mol = -135.5 kJ/mol (exergonic) | | | | |

| potassium hydroxide | manganese dioxide | potassium chlorate | water | potassium permanganate | potassium chloride Hill formula | HKO | MnO_2 | ClKO_3 | H_2O | KMnO_4 | ClK name | potassium hydroxide | manganese dioxide | potassium chlorate | water | potassium permanganate | potassium chloride IUPAC name | potassium hydroxide | dioxomanganese | potassium chlorate | water | potassium permanganate | potassium chloride