Fe(OH)3 + H2SO3 = H2O + FeH(SO3)2

Fe(OH)_3 iron(III) hydroxide + H_2SO_3 sulfurous acid ⟶ H_2O water + FeH(SO3)2

Balance the chemical equation algebraically: Fe(OH)_3 + H_2SO_3 ⟶ H_2O + FeH(SO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe(OH)_3 + c_2 H_2SO_3 ⟶ c_3 H_2O + c_4 FeH(SO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, H, O and S: Fe: | c_1 = c_4 H: | 3 c_1 + 2 c_2 = 2 c_3 + c_4 O: | 3 c_1 + 3 c_2 = c_3 + 6 c_4 S: | c_2 = 2 c_4 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 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Fe(OH)_3 + 2 H_2SO_3 ⟶ 3 H_2O + FeH(SO3)2