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H3PO4 + Fe(OH)2 = H2O + Fe3(PO4)2

Input interpretation

H_3PO_4 phosphoric acid + Fe(OH)_2 iron(II) hydroxide ⟶ H_2O water + Fe3(PO4)2
H_3PO_4 phosphoric acid + Fe(OH)_2 iron(II) hydroxide ⟶ H_2O water + Fe3(PO4)2

Balanced equation

Balance the chemical equation algebraically: H_3PO_4 + Fe(OH)_2 ⟶ H_2O + Fe3(PO4)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_3PO_4 + c_2 Fe(OH)_2 ⟶ c_3 H_2O + c_4 Fe3(PO4)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and Fe: H: | 3 c_1 + 2 c_2 = 2 c_3 O: | 4 c_1 + 2 c_2 = c_3 + 8 c_4 P: | c_1 = 2 c_4 Fe: | c_2 = 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 6 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_3PO_4 + 3 Fe(OH)_2 ⟶ 6 H_2O + Fe3(PO4)2
Balance the chemical equation algebraically: H_3PO_4 + Fe(OH)_2 ⟶ H_2O + Fe3(PO4)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_3PO_4 + c_2 Fe(OH)_2 ⟶ c_3 H_2O + c_4 Fe3(PO4)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and Fe: H: | 3 c_1 + 2 c_2 = 2 c_3 O: | 4 c_1 + 2 c_2 = c_3 + 8 c_4 P: | c_1 = 2 c_4 Fe: | c_2 = 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 6 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_3PO_4 + 3 Fe(OH)_2 ⟶ 6 H_2O + Fe3(PO4)2

Structures

+ ⟶ + Fe3(PO4)2
+ ⟶ + Fe3(PO4)2

Names

phosphoric acid + iron(II) hydroxide ⟶ water + Fe3(PO4)2
phosphoric acid + iron(II) hydroxide ⟶ water + Fe3(PO4)2

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_3PO_4 + Fe(OH)_2 ⟶ H_2O + Fe3(PO4)2 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: 2 H_3PO_4 + 3 Fe(OH)_2 ⟶ 6 H_2O + Fe3(PO4)2 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 H_3PO_4 | 2 | -2 Fe(OH)_2 | 3 | -3 H_2O | 6 | 6 Fe3(PO4)2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_3PO_4 | 2 | -2 | ([H3PO4])^(-2) Fe(OH)_2 | 3 | -3 | ([Fe(OH)2])^(-3) H_2O | 6 | 6 | ([H2O])^6 Fe3(PO4)2 | 1 | 1 | [Fe3(PO4)2] 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 = ([H3PO4])^(-2) ([Fe(OH)2])^(-3) ([H2O])^6 [Fe3(PO4)2] = (([H2O])^6 [Fe3(PO4)2])/(([H3PO4])^2 ([Fe(OH)2])^3)
Construct the equilibrium constant, K, expression for: H_3PO_4 + Fe(OH)_2 ⟶ H_2O + Fe3(PO4)2 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: 2 H_3PO_4 + 3 Fe(OH)_2 ⟶ 6 H_2O + Fe3(PO4)2 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 H_3PO_4 | 2 | -2 Fe(OH)_2 | 3 | -3 H_2O | 6 | 6 Fe3(PO4)2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_3PO_4 | 2 | -2 | ([H3PO4])^(-2) Fe(OH)_2 | 3 | -3 | ([Fe(OH)2])^(-3) H_2O | 6 | 6 | ([H2O])^6 Fe3(PO4)2 | 1 | 1 | [Fe3(PO4)2] 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 = ([H3PO4])^(-2) ([Fe(OH)2])^(-3) ([H2O])^6 [Fe3(PO4)2] = (([H2O])^6 [Fe3(PO4)2])/(([H3PO4])^2 ([Fe(OH)2])^3)

Rate of reaction

Construct the rate of reaction expression for: H_3PO_4 + Fe(OH)_2 ⟶ H_2O + Fe3(PO4)2 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: 2 H_3PO_4 + 3 Fe(OH)_2 ⟶ 6 H_2O + Fe3(PO4)2 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 H_3PO_4 | 2 | -2 Fe(OH)_2 | 3 | -3 H_2O | 6 | 6 Fe3(PO4)2 | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term H_3PO_4 | 2 | -2 | -1/2 (Δ[H3PO4])/(Δt) Fe(OH)_2 | 3 | -3 | -1/3 (Δ[Fe(OH)2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) Fe3(PO4)2 | 1 | 1 | (Δ[Fe3(PO4)2])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -1/2 (Δ[H3PO4])/(Δt) = -1/3 (Δ[Fe(OH)2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[Fe3(PO4)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_3PO_4 + Fe(OH)_2 ⟶ H_2O + Fe3(PO4)2 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: 2 H_3PO_4 + 3 Fe(OH)_2 ⟶ 6 H_2O + Fe3(PO4)2 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 H_3PO_4 | 2 | -2 Fe(OH)_2 | 3 | -3 H_2O | 6 | 6 Fe3(PO4)2 | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term H_3PO_4 | 2 | -2 | -1/2 (Δ[H3PO4])/(Δt) Fe(OH)_2 | 3 | -3 | -1/3 (Δ[Fe(OH)2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) Fe3(PO4)2 | 1 | 1 | (Δ[Fe3(PO4)2])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -1/2 (Δ[H3PO4])/(Δt) = -1/3 (Δ[Fe(OH)2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[Fe3(PO4)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| phosphoric acid | iron(II) hydroxide | water | Fe3(PO4)2 formula | H_3PO_4 | Fe(OH)_2 | H_2O | Fe3(PO4)2 Hill formula | H_3O_4P | FeH_2O_2 | H_2O | Fe3O8P2 name | phosphoric acid | iron(II) hydroxide | water | IUPAC name | phosphoric acid | ferrous dihydroxide | water |
| phosphoric acid | iron(II) hydroxide | water | Fe3(PO4)2 formula | H_3PO_4 | Fe(OH)_2 | H_2O | Fe3(PO4)2 Hill formula | H_3O_4P | FeH_2O_2 | H_2O | Fe3O8P2 name | phosphoric acid | iron(II) hydroxide | water | IUPAC name | phosphoric acid | ferrous dihydroxide | water |

Substance properties

| phosphoric acid | iron(II) hydroxide | water | Fe3(PO4)2 molar mass | 97.994 g/mol | 89.86 g/mol | 18.015 g/mol | 357.47 g/mol phase | liquid (at STP) | | liquid (at STP) | melting point | 42.4 °C | | 0 °C | boiling point | 158 °C | | 99.9839 °C | density | 1.685 g/cm^3 | | 1 g/cm^3 | solubility in water | very soluble | | | surface tension | | | 0.0728 N/m | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | | odorless |
| phosphoric acid | iron(II) hydroxide | water | Fe3(PO4)2 molar mass | 97.994 g/mol | 89.86 g/mol | 18.015 g/mol | 357.47 g/mol phase | liquid (at STP) | | liquid (at STP) | melting point | 42.4 °C | | 0 °C | boiling point | 158 °C | | 99.9839 °C | density | 1.685 g/cm^3 | | 1 g/cm^3 | solubility in water | very soluble | | | surface tension | | | 0.0728 N/m | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | | odorless |

Units

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