H2O + AgNO3 + AsH3 = HNO3 + Ag + H3AsO4

water + silver nitrate + arsine ⟶ nitric acid + silver + arsenic acid, solid

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, Ag, N and As: H: | 2 c_1 + 3 c_3 = c_4 + 3 c_6 O: | c_1 + 3 c_2 = 3 c_4 + 4 c_6 Ag: | c_2 = c_5 N: | c_2 = c_4 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 = 4 c_2 = 8 c_3 = 1 c_4 = 8 c_5 = 8 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 + 8 + ⟶ 8 + 8 +

+ + ⟶ + +

water + silver nitrate + arsine ⟶ nitric acid + silver + arsenic acid, solid

K_c = ([HNO3]^8 [Ag]^8 [H3AsO4])/([H2O]^4 [AgNO3]^8 [AsH3])

rate = -1/4 (Δ[H2O])/(Δt) = -1/8 (Δ[AgNO3])/(Δt) = -(Δ[AsH3])/(Δt) = 1/8 (Δ[HNO3])/(Δt) = 1/8 (Δ[Ag])/(Δt) = (Δ[H3AsO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | silver nitrate | arsine | nitric acid | silver | arsenic acid, solid Hill formula | H_2O | AgNO_3 | AsH_3 | HNO_3 | Ag | AsH_3O_4 name | water | silver nitrate | arsine | nitric acid | silver | arsenic acid, solid IUPAC name | water | silver nitrate | arsane | nitric acid | silver | arsoric acid