Search

HNO3 + FeI2 = H2O + NO2 + HIO3 + Fe(NO3)2

Input interpretation

HNO_3 nitric acid + FeI_2 ferrous iodide ⟶ H_2O water + NO_2 nitrogen dioxide + HIO_3 iodic acid + Fe(NO_3)_2 iron(II) nitrate
HNO_3 nitric acid + FeI_2 ferrous iodide ⟶ H_2O water + NO_2 nitrogen dioxide + HIO_3 iodic acid + Fe(NO_3)_2 iron(II) nitrate

Balanced equation

Balance the chemical equation algebraically: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + HIO_3 + Fe(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 FeI_2 ⟶ c_3 H_2O + c_4 NO_2 + c_5 HIO_3 + c_6 Fe(NO_3)_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Fe and I: H: | c_1 = 2 c_3 + c_5 N: | c_1 = c_4 + 2 c_6 O: | 3 c_1 = c_3 + 2 c_4 + 3 c_5 + 6 c_6 Fe: | c_2 = c_6 I: | 2 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 14 c_2 = 1 c_3 = 6 c_4 = 12 c_5 = 2 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HNO_3 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + Fe(NO_3)_2
Balance the chemical equation algebraically: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + HIO_3 + Fe(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 FeI_2 ⟶ c_3 H_2O + c_4 NO_2 + c_5 HIO_3 + c_6 Fe(NO_3)_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Fe and I: H: | c_1 = 2 c_3 + c_5 N: | c_1 = c_4 + 2 c_6 O: | 3 c_1 = c_3 + 2 c_4 + 3 c_5 + 6 c_6 Fe: | c_2 = c_6 I: | 2 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 14 c_2 = 1 c_3 = 6 c_4 = 12 c_5 = 2 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HNO_3 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + Fe(NO_3)_2

Structures

+ ⟶ + + +
+ ⟶ + + +

Names

nitric acid + ferrous iodide ⟶ water + nitrogen dioxide + iodic acid + iron(II) nitrate
nitric acid + ferrous iodide ⟶ water + nitrogen dioxide + iodic acid + iron(II) nitrate

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + HIO_3 + Fe(NO_3)_2 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 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + Fe(NO_3)_2 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 FeI_2 | 1 | -1 H_2O | 6 | 6 NO_2 | 12 | 12 HIO_3 | 2 | 2 Fe(NO_3)_2 | 1 | 1 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) FeI_2 | 1 | -1 | ([FeI2])^(-1) H_2O | 6 | 6 | ([H2O])^6 NO_2 | 12 | 12 | ([NO2])^12 HIO_3 | 2 | 2 | ([HIO3])^2 Fe(NO_3)_2 | 1 | 1 | [Fe(NO3)2] 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) ([FeI2])^(-1) ([H2O])^6 ([NO2])^12 ([HIO3])^2 [Fe(NO3)2] = (([H2O])^6 ([NO2])^12 ([HIO3])^2 [Fe(NO3)2])/(([HNO3])^14 [FeI2])
Construct the equilibrium constant, K, expression for: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + HIO_3 + Fe(NO_3)_2 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 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + Fe(NO_3)_2 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 FeI_2 | 1 | -1 H_2O | 6 | 6 NO_2 | 12 | 12 HIO_3 | 2 | 2 Fe(NO_3)_2 | 1 | 1 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) FeI_2 | 1 | -1 | ([FeI2])^(-1) H_2O | 6 | 6 | ([H2O])^6 NO_2 | 12 | 12 | ([NO2])^12 HIO_3 | 2 | 2 | ([HIO3])^2 Fe(NO_3)_2 | 1 | 1 | [Fe(NO3)2] 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) ([FeI2])^(-1) ([H2O])^6 ([NO2])^12 ([HIO3])^2 [Fe(NO3)2] = (([H2O])^6 ([NO2])^12 ([HIO3])^2 [Fe(NO3)2])/(([HNO3])^14 [FeI2])

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + HIO_3 + Fe(NO_3)_2 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 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + Fe(NO_3)_2 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 FeI_2 | 1 | -1 H_2O | 6 | 6 NO_2 | 12 | 12 HIO_3 | 2 | 2 Fe(NO_3)_2 | 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 HNO_3 | 14 | -14 | -1/14 (Δ[HNO3])/(Δt) FeI_2 | 1 | -1 | -(Δ[FeI2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) NO_2 | 12 | 12 | 1/12 (Δ[NO2])/(Δt) HIO_3 | 2 | 2 | 1/2 (Δ[HIO3])/(Δt) Fe(NO_3)_2 | 1 | 1 | (Δ[Fe(NO3)2])/(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) = -(Δ[FeI2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/12 (Δ[NO2])/(Δt) = 1/2 (Δ[HIO3])/(Δt) = (Δ[Fe(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + HIO_3 + Fe(NO_3)_2 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 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + Fe(NO_3)_2 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 FeI_2 | 1 | -1 H_2O | 6 | 6 NO_2 | 12 | 12 HIO_3 | 2 | 2 Fe(NO_3)_2 | 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 HNO_3 | 14 | -14 | -1/14 (Δ[HNO3])/(Δt) FeI_2 | 1 | -1 | -(Δ[FeI2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) NO_2 | 12 | 12 | 1/12 (Δ[NO2])/(Δt) HIO_3 | 2 | 2 | 1/2 (Δ[HIO3])/(Δt) Fe(NO_3)_2 | 1 | 1 | (Δ[Fe(NO3)2])/(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) = -(Δ[FeI2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/12 (Δ[NO2])/(Δt) = 1/2 (Δ[HIO3])/(Δt) = (Δ[Fe(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | ferrous iodide | water | nitrogen dioxide | iodic acid | iron(II) nitrate formula | HNO_3 | FeI_2 | H_2O | NO_2 | HIO_3 | Fe(NO_3)_2 Hill formula | HNO_3 | FeI_2 | H_2O | NO_2 | HIO_3 | FeN_2O_6 name | nitric acid | ferrous iodide | water | nitrogen dioxide | iodic acid | iron(II) nitrate IUPAC name | nitric acid | diiodoiron | water | Nitrogen dioxide | iodic acid |
| nitric acid | ferrous iodide | water | nitrogen dioxide | iodic acid | iron(II) nitrate formula | HNO_3 | FeI_2 | H_2O | NO_2 | HIO_3 | Fe(NO_3)_2 Hill formula | HNO_3 | FeI_2 | H_2O | NO_2 | HIO_3 | FeN_2O_6 name | nitric acid | ferrous iodide | water | nitrogen dioxide | iodic acid | iron(II) nitrate IUPAC name | nitric acid | diiodoiron | water | Nitrogen dioxide | iodic acid |

Random Queries

chemical types of 1-methylpyrene vs poly(ethylene-co-1-butene)
alternate names of telluride anion vs fluoride anion
density of barium chlorate vs silver perchlorate
chemical types of pyrophosphoric acid vs oxalic acid
total dynamic head
chemical types of sodium periodate trihydrate vs lead(II) cyanamide
alternate names of silver chloride vs tritium deuteride
half-life of hydrogen vs neon
color of scandium vs dysprosium
lead(II) hexafluorosilicate vs gadolinium