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Fe(NO3)3 + CaS = Ca(NO3)2 + Fe2S3

Input interpretation

Fe(NO_3)_3 ferric nitrate + CaS calcium sulfide ⟶ Ca(NO_3)_2 calcium nitrate + Fe2S3
Fe(NO_3)_3 ferric nitrate + CaS calcium sulfide ⟶ Ca(NO_3)_2 calcium nitrate + Fe2S3

Balanced equation

Balance the chemical equation algebraically: Fe(NO_3)_3 + CaS ⟶ Ca(NO_3)_2 + Fe2S3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe(NO_3)_3 + c_2 CaS ⟶ c_3 Ca(NO_3)_2 + c_4 Fe2S3 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, N, O, Ca and S: Fe: | c_1 = 2 c_4 N: | 3 c_1 = 2 c_3 O: | 9 c_1 = 6 c_3 Ca: | c_2 = c_3 S: | 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 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 2 Fe(NO_3)_3 + 3 CaS ⟶ 3 Ca(NO_3)_2 + Fe2S3
Balance the chemical equation algebraically: Fe(NO_3)_3 + CaS ⟶ Ca(NO_3)_2 + Fe2S3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe(NO_3)_3 + c_2 CaS ⟶ c_3 Ca(NO_3)_2 + c_4 Fe2S3 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, N, O, Ca and S: Fe: | c_1 = 2 c_4 N: | 3 c_1 = 2 c_3 O: | 9 c_1 = 6 c_3 Ca: | c_2 = c_3 S: | 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 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Fe(NO_3)_3 + 3 CaS ⟶ 3 Ca(NO_3)_2 + Fe2S3

Structures

 + ⟶ + Fe2S3
+ ⟶ + Fe2S3

Names

ferric nitrate + calcium sulfide ⟶ calcium nitrate + Fe2S3
ferric nitrate + calcium sulfide ⟶ calcium nitrate + Fe2S3

Equilibrium constant

Construct the equilibrium constant, K, expression for: Fe(NO_3)_3 + CaS ⟶ Ca(NO_3)_2 + Fe2S3 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 Fe(NO_3)_3 + 3 CaS ⟶ 3 Ca(NO_3)_2 + Fe2S3 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(NO_3)_3 | 2 | -2 CaS | 3 | -3 Ca(NO_3)_2 | 3 | 3 Fe2S3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Fe(NO_3)_3 | 2 | -2 | ([Fe(NO3)3])^(-2) CaS | 3 | -3 | ([CaS])^(-3) Ca(NO_3)_2 | 3 | 3 | ([Ca(NO3)2])^3 Fe2S3 | 1 | 1 | [Fe2S3] 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 = ([Fe(NO3)3])^(-2) ([CaS])^(-3) ([Ca(NO3)2])^3 [Fe2S3] = (([Ca(NO3)2])^3 [Fe2S3])/(([Fe(NO3)3])^2 ([CaS])^3)
Construct the equilibrium constant, K, expression for: Fe(NO_3)_3 + CaS ⟶ Ca(NO_3)_2 + Fe2S3 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 Fe(NO_3)_3 + 3 CaS ⟶ 3 Ca(NO_3)_2 + Fe2S3 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(NO_3)_3 | 2 | -2 CaS | 3 | -3 Ca(NO_3)_2 | 3 | 3 Fe2S3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Fe(NO_3)_3 | 2 | -2 | ([Fe(NO3)3])^(-2) CaS | 3 | -3 | ([CaS])^(-3) Ca(NO_3)_2 | 3 | 3 | ([Ca(NO3)2])^3 Fe2S3 | 1 | 1 | [Fe2S3] 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 = ([Fe(NO3)3])^(-2) ([CaS])^(-3) ([Ca(NO3)2])^3 [Fe2S3] = (([Ca(NO3)2])^3 [Fe2S3])/(([Fe(NO3)3])^2 ([CaS])^3)

Rate of reaction

Construct the rate of reaction expression for: Fe(NO_3)_3 + CaS ⟶ Ca(NO_3)_2 + Fe2S3 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 Fe(NO_3)_3 + 3 CaS ⟶ 3 Ca(NO_3)_2 + Fe2S3 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(NO_3)_3 | 2 | -2 CaS | 3 | -3 Ca(NO_3)_2 | 3 | 3 Fe2S3 | 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 Fe(NO_3)_3 | 2 | -2 | -1/2 (Δ[Fe(NO3)3])/(Δt) CaS | 3 | -3 | -1/3 (Δ[CaS])/(Δt) Ca(NO_3)_2 | 3 | 3 | 1/3 (Δ[Ca(NO3)2])/(Δt) Fe2S3 | 1 | 1 | (Δ[Fe2S3])/(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[Fe(NO3)3])/(Δt) = -1/3 (Δ[CaS])/(Δt) = 1/3 (Δ[Ca(NO3)2])/(Δt) = (Δ[Fe2S3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: Fe(NO_3)_3 + CaS ⟶ Ca(NO_3)_2 + Fe2S3 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 Fe(NO_3)_3 + 3 CaS ⟶ 3 Ca(NO_3)_2 + Fe2S3 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(NO_3)_3 | 2 | -2 CaS | 3 | -3 Ca(NO_3)_2 | 3 | 3 Fe2S3 | 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 Fe(NO_3)_3 | 2 | -2 | -1/2 (Δ[Fe(NO3)3])/(Δt) CaS | 3 | -3 | -1/3 (Δ[CaS])/(Δt) Ca(NO_3)_2 | 3 | 3 | 1/3 (Δ[Ca(NO3)2])/(Δt) Fe2S3 | 1 | 1 | (Δ[Fe2S3])/(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[Fe(NO3)3])/(Δt) = -1/3 (Δ[CaS])/(Δt) = 1/3 (Δ[Ca(NO3)2])/(Δt) = (Δ[Fe2S3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | ferric nitrate | calcium sulfide | calcium nitrate | Fe2S3 formula | Fe(NO_3)_3 | CaS | Ca(NO_3)_2 | Fe2S3 Hill formula | FeN_3O_9 | CaS | CaN_2O_6 | Fe2S3 name | ferric nitrate | calcium sulfide | calcium nitrate |  IUPAC name | iron(+3) cation trinitrate | thioxocalcium | calcium dinitrate |
| ferric nitrate | calcium sulfide | calcium nitrate | Fe2S3 formula | Fe(NO_3)_3 | CaS | Ca(NO_3)_2 | Fe2S3 Hill formula | FeN_3O_9 | CaS | CaN_2O_6 | Fe2S3 name | ferric nitrate | calcium sulfide | calcium nitrate | IUPAC name | iron(+3) cation trinitrate | thioxocalcium | calcium dinitrate |

Substance properties

 | ferric nitrate | calcium sulfide | calcium nitrate | Fe2S3 molar mass | 241.86 g/mol | 72.14 g/mol | 164.09 g/mol | 207.9 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) |  melting point | 35 °C | 2450 °C | 562 °C |  density | 1.7 g/cm^3 | 2.5 g/cm^3 | 2.5 g/cm^3 |  solubility in water | very soluble | decomposes | soluble |
| ferric nitrate | calcium sulfide | calcium nitrate | Fe2S3 molar mass | 241.86 g/mol | 72.14 g/mol | 164.09 g/mol | 207.9 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | melting point | 35 °C | 2450 °C | 562 °C | density | 1.7 g/cm^3 | 2.5 g/cm^3 | 2.5 g/cm^3 | solubility in water | very soluble | decomposes | soluble |

Units