HNO3 + H3PO4 + (NH4)2MoO4 = H2O + NH4NO3 + (NH4)3PO4 + MoO3

HNO_3 nitric acid + H_3PO_4 phosphoric acid + (NH_4)_2MoO_4 ammonium molybdate ⟶ H_2O water + NH_4NO_3 ammonium nitrate + (NH4)3PO4 + MoO_3 molybdenum trioxide

Balance the chemical equation algebraically: HNO_3 + H_3PO_4 + (NH_4)_2MoO_4 ⟶ H_2O + NH_4NO_3 + (NH4)3PO4 + MoO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 H_3PO_4 + c_3 (NH_4)_2MoO_4 ⟶ c_4 H_2O + c_5 NH_4NO_3 + c_6 (NH4)3PO4 + c_7 MoO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, P and Mo: H: | c_1 + 3 c_2 + 5 c_3 = 2 c_4 + 4 c_5 + 12 c_6 N: | c_1 + c_3 = 2 c_5 + 3 c_6 O: | 3 c_1 + 4 c_2 + 4 c_3 = c_4 + 3 c_5 + 4 c_6 + 3 c_7 P: | c_2 = c_6 Mo: | c_3 = c_7 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_3 = 3 c_2 + 1 c_4 = 3 c_2 + 1 c_5 = 1 c_6 = c_2 c_7 = 3 c_2 + 1 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_2 = 1 and solve for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 4 c_4 = 4 c_5 = 1 c_6 = 1 c_7 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | HNO_3 + H_3PO_4 + 4 (NH_4)_2MoO_4 ⟶ 4 H_2O + NH_4NO_3 + (NH4)3PO4 + 4 MoO_3

+ + ⟶ + + (NH4)3PO4 +

nitric acid + phosphoric acid + ammonium molybdate ⟶ water + ammonium nitrate + (NH4)3PO4 + molybdenum trioxide

Construct the equilibrium constant, K, expression for: HNO_3 + H_3PO_4 + (NH_4)_2MoO_4 ⟶ H_2O + NH_4NO_3 + (NH4)3PO4 + MoO_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: HNO_3 + H_3PO_4 + 4 (NH_4)_2MoO_4 ⟶ 4 H_2O + NH_4NO_3 + (NH4)3PO4 + 4 MoO_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HNO_3 | 1 | -1 H_3PO_4 | 1 | -1 (NH_4)_2MoO_4 | 4 | -4 H_2O | 4 | 4 NH_4NO_3 | 1 | 1 (NH4)3PO4 | 1 | 1 MoO_3 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 1 | -1 | ([HNO3])^(-1) H_3PO_4 | 1 | -1 | ([H3PO4])^(-1) (NH_4)_2MoO_4 | 4 | -4 | ([(NH4)2MoO4])^(-4) H_2O | 4 | 4 | ([H2O])^4 NH_4NO_3 | 1 | 1 | [NH4NO3] (NH4)3PO4 | 1 | 1 | [(NH4)3PO4] MoO_3 | 4 | 4 | ([MoO3])^4 The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([HNO3])^(-1) ([H3PO4])^(-1) ([(NH4)2MoO4])^(-4) ([H2O])^4 [NH4NO3] [(NH4)3PO4] ([MoO3])^4 = (([H2O])^4 [NH4NO3] [(NH4)3PO4] ([MoO3])^4)/([HNO3] [H3PO4] ([(NH4)2MoO4])^4)