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NaOH + Fe(OH)3 = H2O + NaFeO2

Input interpretation

NaOH sodium hydroxide + Fe(OH)_3 iron(III) hydroxide ⟶ H_2O water + NaFeO2
NaOH sodium hydroxide + Fe(OH)_3 iron(III) hydroxide ⟶ H_2O water + NaFeO2

Balanced equation

Balance the chemical equation algebraically: NaOH + Fe(OH)_3 ⟶ H_2O + NaFeO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Fe(OH)_3 ⟶ c_3 H_2O + c_4 NaFeO2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O and Fe: H: | c_1 + 3 c_2 = 2 c_3 Na: | c_1 = c_4 O: | c_1 + 3 c_2 = c_3 + 2 c_4 Fe: | c_2 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | NaOH + Fe(OH)_3 ⟶ 2 H_2O + NaFeO2
Balance the chemical equation algebraically: NaOH + Fe(OH)_3 ⟶ H_2O + NaFeO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Fe(OH)_3 ⟶ c_3 H_2O + c_4 NaFeO2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O and Fe: H: | c_1 + 3 c_2 = 2 c_3 Na: | c_1 = c_4 O: | c_1 + 3 c_2 = c_3 + 2 c_4 Fe: | c_2 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | NaOH + Fe(OH)_3 ⟶ 2 H_2O + NaFeO2

Structures

+ ⟶ + NaFeO2
+ ⟶ + NaFeO2

Names

sodium hydroxide + iron(III) hydroxide ⟶ water + NaFeO2
sodium hydroxide + iron(III) hydroxide ⟶ water + NaFeO2

Equilibrium constant

Construct the equilibrium constant, K, expression for: NaOH + Fe(OH)_3 ⟶ H_2O + NaFeO2 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: NaOH + Fe(OH)_3 ⟶ 2 H_2O + NaFeO2 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 NaOH | 1 | -1 Fe(OH)_3 | 1 | -1 H_2O | 2 | 2 NaFeO2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 1 | -1 | ([NaOH])^(-1) Fe(OH)_3 | 1 | -1 | ([Fe(OH)3])^(-1) H_2O | 2 | 2 | ([H2O])^2 NaFeO2 | 1 | 1 | [NaFeO2] 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 = ([NaOH])^(-1) ([Fe(OH)3])^(-1) ([H2O])^2 [NaFeO2] = (([H2O])^2 [NaFeO2])/([NaOH] [Fe(OH)3])
Construct the equilibrium constant, K, expression for: NaOH + Fe(OH)_3 ⟶ H_2O + NaFeO2 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: NaOH + Fe(OH)_3 ⟶ 2 H_2O + NaFeO2 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 NaOH | 1 | -1 Fe(OH)_3 | 1 | -1 H_2O | 2 | 2 NaFeO2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 1 | -1 | ([NaOH])^(-1) Fe(OH)_3 | 1 | -1 | ([Fe(OH)3])^(-1) H_2O | 2 | 2 | ([H2O])^2 NaFeO2 | 1 | 1 | [NaFeO2] 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 = ([NaOH])^(-1) ([Fe(OH)3])^(-1) ([H2O])^2 [NaFeO2] = (([H2O])^2 [NaFeO2])/([NaOH] [Fe(OH)3])

Rate of reaction

Construct the rate of reaction expression for: NaOH + Fe(OH)_3 ⟶ H_2O + NaFeO2 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: NaOH + Fe(OH)_3 ⟶ 2 H_2O + NaFeO2 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 NaOH | 1 | -1 Fe(OH)_3 | 1 | -1 H_2O | 2 | 2 NaFeO2 | 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 NaOH | 1 | -1 | -(Δ[NaOH])/(Δt) Fe(OH)_3 | 1 | -1 | -(Δ[Fe(OH)3])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) NaFeO2 | 1 | 1 | (Δ[NaFeO2])/(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 = -(Δ[NaOH])/(Δt) = -(Δ[Fe(OH)3])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[NaFeO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: NaOH + Fe(OH)_3 ⟶ H_2O + NaFeO2 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: NaOH + Fe(OH)_3 ⟶ 2 H_2O + NaFeO2 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 NaOH | 1 | -1 Fe(OH)_3 | 1 | -1 H_2O | 2 | 2 NaFeO2 | 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 NaOH | 1 | -1 | -(Δ[NaOH])/(Δt) Fe(OH)_3 | 1 | -1 | -(Δ[Fe(OH)3])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) NaFeO2 | 1 | 1 | (Δ[NaFeO2])/(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 = -(Δ[NaOH])/(Δt) = -(Δ[Fe(OH)3])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[NaFeO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| sodium hydroxide | iron(III) hydroxide | water | NaFeO2 formula | NaOH | Fe(OH)_3 | H_2O | NaFeO2 Hill formula | HNaO | FeH_3O_3 | H_2O | FeNaO2 name | sodium hydroxide | iron(III) hydroxide | water | IUPAC name | sodium hydroxide | ferric trihydroxide | water |
| sodium hydroxide | iron(III) hydroxide | water | NaFeO2 formula | NaOH | Fe(OH)_3 | H_2O | NaFeO2 Hill formula | HNaO | FeH_3O_3 | H_2O | FeNaO2 name | sodium hydroxide | iron(III) hydroxide | water | IUPAC name | sodium hydroxide | ferric trihydroxide | water |

Substance properties

| sodium hydroxide | iron(III) hydroxide | water | NaFeO2 molar mass | 39.997 g/mol | 106.87 g/mol | 18.015 g/mol | 110.83 g/mol phase | solid (at STP) | | liquid (at STP) | melting point | 323 °C | | 0 °C | boiling point | 1390 °C | | 99.9839 °C | density | 2.13 g/cm^3 | | 1 g/cm^3 | solubility in water | soluble | | | surface tension | 0.07435 N/m | | 0.0728 N/m | dynamic viscosity | 0.004 Pa s (at 350 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | | | odorless |
| sodium hydroxide | iron(III) hydroxide | water | NaFeO2 molar mass | 39.997 g/mol | 106.87 g/mol | 18.015 g/mol | 110.83 g/mol phase | solid (at STP) | | liquid (at STP) | melting point | 323 °C | | 0 °C | boiling point | 1390 °C | | 99.9839 °C | density | 2.13 g/cm^3 | | 1 g/cm^3 | solubility in water | soluble | | | surface tension | 0.07435 N/m | | 0.0728 N/m | dynamic viscosity | 0.004 Pa s (at 350 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | | | odorless |

Units

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