NaOH + NaClO + As = H2O + NaCl + Na3AsO4

sodium hydroxide + sodium hypochlorite + gray arsenic ⟶ water + sodium chloride + arsenic acid, trisodium salt

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, Na, O, Cl and As: H: | c_1 = 2 c_4 Na: | c_1 + c_2 = c_5 + 3 c_6 O: | c_1 + c_2 = c_4 + 4 c_6 Cl: | c_2 = c_5 As: | 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 = 3 c_2 = 5/2 c_3 = 1 c_4 = 3/2 c_5 = 5/2 c_6 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 6 c_2 = 5 c_3 = 2 c_4 = 3 c_5 = 5 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 + 5 + 2 ⟶ 3 + 5 + 2

+ + ⟶ + +

sodium hydroxide + sodium hypochlorite + gray arsenic ⟶ water + sodium chloride + arsenic acid, trisodium salt

K_c = ([H2O]^3 [NaCl]^5 [AsNa3O4]^2)/([NaOH]^6 [NaOCl]^5 [As]^2)

rate = -1/6 (Δ[NaOH])/(Δt) = -1/5 (Δ[NaOCl])/(Δt) = -1/2 (Δ[As])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/5 (Δ[NaCl])/(Δt) = 1/2 (Δ[AsNa3O4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium hydroxide | sodium hypochlorite | gray arsenic | water | sodium chloride | arsenic acid, trisodium salt Hill formula | HNaO | ClNaO | As | H_2O | ClNa | AsNa_3O_4 name | sodium hydroxide | sodium hypochlorite | gray arsenic | water | sodium chloride | arsenic acid, trisodium salt IUPAC name | sodium hydroxide | sodium hypochlorite | arsenic | water | sodium chloride |