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Fe2(SO4)3 + LiOH = Fe(OH)3 + Li2SO4

Input interpretation

Fe_2(SO_4)_3·xH_2O iron(III) sulfate hydrate + LiOH lithium hydroxide ⟶ Fe(OH)_3 iron(III) hydroxide + Li_2SO_4 lithium sulfate
Fe_2(SO_4)_3·xH_2O iron(III) sulfate hydrate + LiOH lithium hydroxide ⟶ Fe(OH)_3 iron(III) hydroxide + Li_2SO_4 lithium sulfate

Balanced equation

Balance the chemical equation algebraically: Fe_2(SO_4)_3·xH_2O + LiOH ⟶ Fe(OH)_3 + Li_2SO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe_2(SO_4)_3·xH_2O + c_2 LiOH ⟶ c_3 Fe(OH)_3 + c_4 Li_2SO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, O, S, H and Li: Fe: | 2 c_1 = c_3 O: | 12 c_1 + c_2 = 3 c_3 + 4 c_4 S: | 3 c_1 = c_4 H: | c_2 = 3 c_3 Li: | c_2 = 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 = 6 c_3 = 2 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Fe_2(SO_4)_3·xH_2O + 6 LiOH ⟶ 2 Fe(OH)_3 + 3 Li_2SO_4
Balance the chemical equation algebraically: Fe_2(SO_4)_3·xH_2O + LiOH ⟶ Fe(OH)_3 + Li_2SO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe_2(SO_4)_3·xH_2O + c_2 LiOH ⟶ c_3 Fe(OH)_3 + c_4 Li_2SO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, O, S, H and Li: Fe: | 2 c_1 = c_3 O: | 12 c_1 + c_2 = 3 c_3 + 4 c_4 S: | 3 c_1 = c_4 H: | c_2 = 3 c_3 Li: | c_2 = 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 = 6 c_3 = 2 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Fe_2(SO_4)_3·xH_2O + 6 LiOH ⟶ 2 Fe(OH)_3 + 3 Li_2SO_4

Structures

+ ⟶ +
+ ⟶ +

Names

iron(III) sulfate hydrate + lithium hydroxide ⟶ iron(III) hydroxide + lithium sulfate
iron(III) sulfate hydrate + lithium hydroxide ⟶ iron(III) hydroxide + lithium sulfate

Equilibrium constant

Construct the equilibrium constant, K, expression for: Fe_2(SO_4)_3·xH_2O + LiOH ⟶ Fe(OH)_3 + Li_2SO_4 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: Fe_2(SO_4)_3·xH_2O + 6 LiOH ⟶ 2 Fe(OH)_3 + 3 Li_2SO_4 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 Fe_2(SO_4)_3·xH_2O | 1 | -1 LiOH | 6 | -6 Fe(OH)_3 | 2 | 2 Li_2SO_4 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Fe_2(SO_4)_3·xH_2O | 1 | -1 | ([Fe2(SO4)3·xH2O])^(-1) LiOH | 6 | -6 | ([LiOH])^(-6) Fe(OH)_3 | 2 | 2 | ([Fe(OH)3])^2 Li_2SO_4 | 3 | 3 | ([Li2SO4])^3 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 = ([Fe2(SO4)3·xH2O])^(-1) ([LiOH])^(-6) ([Fe(OH)3])^2 ([Li2SO4])^3 = (([Fe(OH)3])^2 ([Li2SO4])^3)/([Fe2(SO4)3·xH2O] ([LiOH])^6)
Construct the equilibrium constant, K, expression for: Fe_2(SO_4)_3·xH_2O + LiOH ⟶ Fe(OH)_3 + Li_2SO_4 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: Fe_2(SO_4)_3·xH_2O + 6 LiOH ⟶ 2 Fe(OH)_3 + 3 Li_2SO_4 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 Fe_2(SO_4)_3·xH_2O | 1 | -1 LiOH | 6 | -6 Fe(OH)_3 | 2 | 2 Li_2SO_4 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Fe_2(SO_4)_3·xH_2O | 1 | -1 | ([Fe2(SO4)3·xH2O])^(-1) LiOH | 6 | -6 | ([LiOH])^(-6) Fe(OH)_3 | 2 | 2 | ([Fe(OH)3])^2 Li_2SO_4 | 3 | 3 | ([Li2SO4])^3 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 = ([Fe2(SO4)3·xH2O])^(-1) ([LiOH])^(-6) ([Fe(OH)3])^2 ([Li2SO4])^3 = (([Fe(OH)3])^2 ([Li2SO4])^3)/([Fe2(SO4)3·xH2O] ([LiOH])^6)

Rate of reaction

Construct the rate of reaction expression for: Fe_2(SO_4)_3·xH_2O + LiOH ⟶ Fe(OH)_3 + Li_2SO_4 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: Fe_2(SO_4)_3·xH_2O + 6 LiOH ⟶ 2 Fe(OH)_3 + 3 Li_2SO_4 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 Fe_2(SO_4)_3·xH_2O | 1 | -1 LiOH | 6 | -6 Fe(OH)_3 | 2 | 2 Li_2SO_4 | 3 | 3 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 Fe_2(SO_4)_3·xH_2O | 1 | -1 | -(Δ[Fe2(SO4)3·xH2O])/(Δt) LiOH | 6 | -6 | -1/6 (Δ[LiOH])/(Δt) Fe(OH)_3 | 2 | 2 | 1/2 (Δ[Fe(OH)3])/(Δt) Li_2SO_4 | 3 | 3 | 1/3 (Δ[Li2SO4])/(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[Fe2(SO4)3·xH2O])/(Δt) = -1/6 (Δ[LiOH])/(Δt) = 1/2 (Δ[Fe(OH)3])/(Δt) = 1/3 (Δ[Li2SO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: Fe_2(SO_4)_3·xH_2O + LiOH ⟶ Fe(OH)_3 + Li_2SO_4 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: Fe_2(SO_4)_3·xH_2O + 6 LiOH ⟶ 2 Fe(OH)_3 + 3 Li_2SO_4 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 Fe_2(SO_4)_3·xH_2O | 1 | -1 LiOH | 6 | -6 Fe(OH)_3 | 2 | 2 Li_2SO_4 | 3 | 3 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 Fe_2(SO_4)_3·xH_2O | 1 | -1 | -(Δ[Fe2(SO4)3·xH2O])/(Δt) LiOH | 6 | -6 | -1/6 (Δ[LiOH])/(Δt) Fe(OH)_3 | 2 | 2 | 1/2 (Δ[Fe(OH)3])/(Δt) Li_2SO_4 | 3 | 3 | 1/3 (Δ[Li2SO4])/(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[Fe2(SO4)3·xH2O])/(Δt) = -1/6 (Δ[LiOH])/(Δt) = 1/2 (Δ[Fe(OH)3])/(Δt) = 1/3 (Δ[Li2SO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| iron(III) sulfate hydrate | lithium hydroxide | iron(III) hydroxide | lithium sulfate formula | Fe_2(SO_4)_3·xH_2O | LiOH | Fe(OH)_3 | Li_2SO_4 Hill formula | Fe_2O_12S_3 | HLiO | FeH_3O_3 | Li_2O_4S name | iron(III) sulfate hydrate | lithium hydroxide | iron(III) hydroxide | lithium sulfate IUPAC name | diferric trisulfate | lithium hydroxide | ferric trihydroxide | dilithium sulfate
| iron(III) sulfate hydrate | lithium hydroxide | iron(III) hydroxide | lithium sulfate formula | Fe_2(SO_4)_3·xH_2O | LiOH | Fe(OH)_3 | Li_2SO_4 Hill formula | Fe_2O_12S_3 | HLiO | FeH_3O_3 | Li_2O_4S name | iron(III) sulfate hydrate | lithium hydroxide | iron(III) hydroxide | lithium sulfate IUPAC name | diferric trisulfate | lithium hydroxide | ferric trihydroxide | dilithium sulfate

Substance properties

| iron(III) sulfate hydrate | lithium hydroxide | iron(III) hydroxide | lithium sulfate molar mass | 399.9 g/mol | 23.95 g/mol | 106.87 g/mol | 109.9 g/mol phase | | solid (at STP) | | solid (at STP) melting point | | 462 °C | | 845 °C boiling point | | | | 1377 °C density | | 1.46 g/cm^3 | | 2.22 g/cm^3 solubility in water | slightly soluble | | | odor | | odorless | |
| iron(III) sulfate hydrate | lithium hydroxide | iron(III) hydroxide | lithium sulfate molar mass | 399.9 g/mol | 23.95 g/mol | 106.87 g/mol | 109.9 g/mol phase | | solid (at STP) | | solid (at STP) melting point | | 462 °C | | 845 °C boiling point | | | | 1377 °C density | | 1.46 g/cm^3 | | 2.22 g/cm^3 solubility in water | slightly soluble | | | odor | | odorless | |

Units

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