H2O + As2O3 + HOCl = HCl + H3AsO4

H_2O water + As_2O_3 arsenic trioxide + HOCl hypochlorous acid ⟶ HCl hydrogen chloride + H_3AsO_4 arsenic acid, solid

Balance the chemical equation algebraically: H_2O + As_2O_3 + HOCl ⟶ HCl + H_3AsO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 As_2O_3 + c_3 HOCl ⟶ c_4 HCl + c_5 H_3AsO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, As and Cl: H: | 2 c_1 + c_3 = c_4 + 3 c_5 O: | c_1 + 3 c_2 + c_3 = 4 c_5 As: | 2 c_2 = c_5 Cl: | c_3 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 1 c_3 = 2 c_4 = 2 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 H_2O + As_2O_3 + 2 HOCl ⟶ 2 HCl + 2 H_3AsO_4

+ + ⟶ +

water + arsenic trioxide + hypochlorous acid ⟶ hydrogen chloride + arsenic acid, solid

K_c = ([HCl]^2 [H3AsO4]^2)/([H2O]^3 [As2O3] [HOCl]^2)

rate = -1/3 (Δ[H2O])/(Δt) = -(Δ[As2O3])/(Δt) = -1/2 (Δ[HOCl])/(Δt) = 1/2 (Δ[HCl])/(Δt) = 1/2 (Δ[H3AsO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | arsenic trioxide | hypochlorous acid | hydrogen chloride | arsenic acid, solid formula | H_2O | As_2O_3 | HOCl | HCl | H_3AsO_4 Hill formula | H_2O | As_2O_3 | ClHO | ClH | AsH_3O_4 name | water | arsenic trioxide | hypochlorous acid | hydrogen chloride | arsenic acid, solid IUPAC name | water | 2, 4, 5-trioxa-1, 3-diarsabicyclo[1.1.1]pentane | hypochlorous acid | hydrogen chloride | arsoric acid