Search

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

Random Queries

(boiling point of xenon)/(critical temperature of xenon)
mass fractions of sodium
density of copper(II) ferrous sulfide vs barium tetracyanoplatinate
melting point of weak acids
structure diagram of titanium(IV) cation vs zirconium(I) cation
images of group 11 elements
formula of cobalt dichloride vs dysprosium(III) carbonate hydrate
Cu + NHO3 = H2O + NO2 + Cu(NO3)2
crystal class of arsenic
molar mass of ferric ammonium oxalate