H2O + KMnO4 + NaNO2 = KOH + MnO2 + NaNO3

water + potassium permanganate + sodium nitrite ⟶ potassium hydroxide + manganese dioxide + sodium nitrate

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, O, K, Mn, N and Na: H: | 2 c_1 = c_4 O: | c_1 + 4 c_2 + 2 c_3 = c_4 + 2 c_5 + 3 c_6 K: | c_2 = c_4 Mn: | c_2 = c_5 N: | c_3 = c_6 Na: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 3 c_4 = 2 c_5 = 2 c_6 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | + 2 + 3 ⟶ 2 + 2 + 3

+ + ⟶ + +

water + potassium permanganate + sodium nitrite ⟶ potassium hydroxide + manganese dioxide + sodium nitrate

| water | potassium permanganate | sodium nitrite | potassium hydroxide | manganese dioxide | sodium nitrate molecular free energy | -237.1 kJ/mol | -737.6 kJ/mol | -284.6 kJ/mol | -379.4 kJ/mol | -465.1 kJ/mol | -366 kJ/mol total free energy | -237.1 kJ/mol | -1475 kJ/mol | -853.8 kJ/mol | -758.8 kJ/mol | -930.2 kJ/mol | -1098 kJ/mol | G_initial = -2566 kJ/mol | | | G_final = -2787 kJ/mol | | ΔG_rxn^0 | -2787 kJ/mol - -2566 kJ/mol = -220.9 kJ/mol (exergonic) | | | | |

| water | potassium permanganate | sodium nitrite | potassium hydroxide | manganese dioxide | sodium nitrate Hill formula | H_2O | KMnO_4 | NNaO_2 | HKO | MnO_2 | NNaO_3 name | water | potassium permanganate | sodium nitrite | potassium hydroxide | manganese dioxide | sodium nitrate IUPAC name | water | potassium permanganate | sodium nitrite | potassium hydroxide | dioxomanganese | sodium nitrate