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HNO3 + FeO = H2O + NO + Fe(NO3)3NO3

Input interpretation

HNO_3 nitric acid + FeO iron(II) oxide ⟶ H_2O water + NO nitric oxide + Fe(NO3)3NO3
HNO_3 nitric acid + FeO iron(II) oxide ⟶ H_2O water + NO nitric oxide + Fe(NO3)3NO3

Balanced equation

Balance the chemical equation algebraically: HNO_3 + FeO ⟶ H_2O + NO + Fe(NO3)3NO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 FeO ⟶ c_3 H_2O + c_4 NO + c_5 Fe(NO3)3NO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Fe: H: | c_1 = 2 c_3 N: | c_1 = c_4 + 4 c_5 O: | 3 c_1 + c_2 = c_3 + c_4 + 12 c_5 Fe: | c_2 = c_5 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 = 7 c_2 = 3/2 c_3 = 7/2 c_4 = 1 c_5 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 14 c_2 = 3 c_3 = 7 c_4 = 2 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HNO_3 + 3 FeO ⟶ 7 H_2O + 2 NO + 3 Fe(NO3)3NO3
Balance the chemical equation algebraically: HNO_3 + FeO ⟶ H_2O + NO + Fe(NO3)3NO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 FeO ⟶ c_3 H_2O + c_4 NO + c_5 Fe(NO3)3NO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Fe: H: | c_1 = 2 c_3 N: | c_1 = c_4 + 4 c_5 O: | 3 c_1 + c_2 = c_3 + c_4 + 12 c_5 Fe: | c_2 = c_5 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 = 7 c_2 = 3/2 c_3 = 7/2 c_4 = 1 c_5 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 14 c_2 = 3 c_3 = 7 c_4 = 2 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HNO_3 + 3 FeO ⟶ 7 H_2O + 2 NO + 3 Fe(NO3)3NO3

Structures

+ ⟶ + + Fe(NO3)3NO3
+ ⟶ + + Fe(NO3)3NO3

Names

nitric acid + iron(II) oxide ⟶ water + nitric oxide + Fe(NO3)3NO3
nitric acid + iron(II) oxide ⟶ water + nitric oxide + Fe(NO3)3NO3

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + FeO ⟶ H_2O + NO + Fe(NO3)3NO3 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: 14 HNO_3 + 3 FeO ⟶ 7 H_2O + 2 NO + 3 Fe(NO3)3NO3 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 HNO_3 | 14 | -14 FeO | 3 | -3 H_2O | 7 | 7 NO | 2 | 2 Fe(NO3)3NO3 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 14 | -14 | ([HNO3])^(-14) FeO | 3 | -3 | ([FeO])^(-3) H_2O | 7 | 7 | ([H2O])^7 NO | 2 | 2 | ([NO])^2 Fe(NO3)3NO3 | 3 | 3 | ([Fe(NO3)3NO3])^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 = ([HNO3])^(-14) ([FeO])^(-3) ([H2O])^7 ([NO])^2 ([Fe(NO3)3NO3])^3 = (([H2O])^7 ([NO])^2 ([Fe(NO3)3NO3])^3)/(([HNO3])^14 ([FeO])^3)
Construct the equilibrium constant, K, expression for: HNO_3 + FeO ⟶ H_2O + NO + Fe(NO3)3NO3 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: 14 HNO_3 + 3 FeO ⟶ 7 H_2O + 2 NO + 3 Fe(NO3)3NO3 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 HNO_3 | 14 | -14 FeO | 3 | -3 H_2O | 7 | 7 NO | 2 | 2 Fe(NO3)3NO3 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 14 | -14 | ([HNO3])^(-14) FeO | 3 | -3 | ([FeO])^(-3) H_2O | 7 | 7 | ([H2O])^7 NO | 2 | 2 | ([NO])^2 Fe(NO3)3NO3 | 3 | 3 | ([Fe(NO3)3NO3])^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 = ([HNO3])^(-14) ([FeO])^(-3) ([H2O])^7 ([NO])^2 ([Fe(NO3)3NO3])^3 = (([H2O])^7 ([NO])^2 ([Fe(NO3)3NO3])^3)/(([HNO3])^14 ([FeO])^3)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + FeO ⟶ H_2O + NO + Fe(NO3)3NO3 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: 14 HNO_3 + 3 FeO ⟶ 7 H_2O + 2 NO + 3 Fe(NO3)3NO3 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 HNO_3 | 14 | -14 FeO | 3 | -3 H_2O | 7 | 7 NO | 2 | 2 Fe(NO3)3NO3 | 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 HNO_3 | 14 | -14 | -1/14 (Δ[HNO3])/(Δt) FeO | 3 | -3 | -1/3 (Δ[FeO])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) NO | 2 | 2 | 1/2 (Δ[NO])/(Δt) Fe(NO3)3NO3 | 3 | 3 | 1/3 (Δ[Fe(NO3)3NO3])/(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 = -1/14 (Δ[HNO3])/(Δt) = -1/3 (Δ[FeO])/(Δt) = 1/7 (Δ[H2O])/(Δt) = 1/2 (Δ[NO])/(Δt) = 1/3 (Δ[Fe(NO3)3NO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + FeO ⟶ H_2O + NO + Fe(NO3)3NO3 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: 14 HNO_3 + 3 FeO ⟶ 7 H_2O + 2 NO + 3 Fe(NO3)3NO3 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 HNO_3 | 14 | -14 FeO | 3 | -3 H_2O | 7 | 7 NO | 2 | 2 Fe(NO3)3NO3 | 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 HNO_3 | 14 | -14 | -1/14 (Δ[HNO3])/(Δt) FeO | 3 | -3 | -1/3 (Δ[FeO])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) NO | 2 | 2 | 1/2 (Δ[NO])/(Δt) Fe(NO3)3NO3 | 3 | 3 | 1/3 (Δ[Fe(NO3)3NO3])/(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 = -1/14 (Δ[HNO3])/(Δt) = -1/3 (Δ[FeO])/(Δt) = 1/7 (Δ[H2O])/(Δt) = 1/2 (Δ[NO])/(Δt) = 1/3 (Δ[Fe(NO3)3NO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | iron(II) oxide | water | nitric oxide | Fe(NO3)3NO3 formula | HNO_3 | FeO | H_2O | NO | Fe(NO3)3NO3 Hill formula | HNO_3 | FeO | H_2O | NO | FeN4O12 name | nitric acid | iron(II) oxide | water | nitric oxide | IUPAC name | nitric acid | oxoiron | water | nitric oxide |
| nitric acid | iron(II) oxide | water | nitric oxide | Fe(NO3)3NO3 formula | HNO_3 | FeO | H_2O | NO | Fe(NO3)3NO3 Hill formula | HNO_3 | FeO | H_2O | NO | FeN4O12 name | nitric acid | iron(II) oxide | water | nitric oxide | IUPAC name | nitric acid | oxoiron | water | nitric oxide |

Substance properties

| nitric acid | iron(II) oxide | water | nitric oxide | Fe(NO3)3NO3 molar mass | 63.012 g/mol | 71.844 g/mol | 18.015 g/mol | 30.006 g/mol | 303.86 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 1360 °C | 0 °C | -163.6 °C | boiling point | 83 °C | | 99.9839 °C | -151.7 °C | density | 1.5129 g/cm^3 | 5.7 g/cm^3 | 1 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | solubility in water | miscible | insoluble | | | surface tension | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.911×10^-5 Pa s (at 25 °C) | odor | | | odorless | |
| nitric acid | iron(II) oxide | water | nitric oxide | Fe(NO3)3NO3 molar mass | 63.012 g/mol | 71.844 g/mol | 18.015 g/mol | 30.006 g/mol | 303.86 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 1360 °C | 0 °C | -163.6 °C | boiling point | 83 °C | | 99.9839 °C | -151.7 °C | density | 1.5129 g/cm^3 | 5.7 g/cm^3 | 1 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | solubility in water | miscible | insoluble | | | surface tension | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.911×10^-5 Pa s (at 25 °C) | odor | | | odorless | |

Units

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