Search

HNO3 + FeI2 = H2O + NO2 + Fe(NO3)3 + HIO3

Input interpretation

HNO_3 nitric acid + FeI_2 ferrous iodide ⟶ H_2O water + NO_2 nitrogen dioxide + Fe(NO_3)_3 ferric nitrate + HIO_3 iodic acid
HNO_3 nitric acid + FeI_2 ferrous iodide ⟶ H_2O water + NO_2 nitrogen dioxide + Fe(NO_3)_3 ferric nitrate + HIO_3 iodic acid

Balanced equation

Balance the chemical equation algebraically: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + Fe(NO_3)_3 + HIO_3 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 Fe(NO_3)_3 + c_6 HIO_3 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_6 N: | c_1 = c_4 + 3 c_5 O: | 3 c_1 = c_3 + 2 c_4 + 9 c_5 + 3 c_6 Fe: | c_2 = c_5 I: | 2 c_2 = c_6 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 = 16 c_2 = 1 c_3 = 7 c_4 = 13 c_5 = 1 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 16 HNO_3 + FeI_2 ⟶ 7 H_2O + 13 NO_2 + Fe(NO_3)_3 + 2 HIO_3
Balance the chemical equation algebraically: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + Fe(NO_3)_3 + HIO_3 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 Fe(NO_3)_3 + c_6 HIO_3 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_6 N: | c_1 = c_4 + 3 c_5 O: | 3 c_1 = c_3 + 2 c_4 + 9 c_5 + 3 c_6 Fe: | c_2 = c_5 I: | 2 c_2 = c_6 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 = 16 c_2 = 1 c_3 = 7 c_4 = 13 c_5 = 1 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 16 HNO_3 + FeI_2 ⟶ 7 H_2O + 13 NO_2 + Fe(NO_3)_3 + 2 HIO_3

Structures

+ ⟶ + + +
+ ⟶ + + +

Names

nitric acid + ferrous iodide ⟶ water + nitrogen dioxide + ferric nitrate + iodic acid
nitric acid + ferrous iodide ⟶ water + nitrogen dioxide + ferric nitrate + iodic acid

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + Fe(NO_3)_3 + HIO_3 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: 16 HNO_3 + FeI_2 ⟶ 7 H_2O + 13 NO_2 + Fe(NO_3)_3 + 2 HIO_3 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 | 16 | -16 FeI_2 | 1 | -1 H_2O | 7 | 7 NO_2 | 13 | 13 Fe(NO_3)_3 | 1 | 1 HIO_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 16 | -16 | ([HNO3])^(-16) FeI_2 | 1 | -1 | ([FeI2])^(-1) H_2O | 7 | 7 | ([H2O])^7 NO_2 | 13 | 13 | ([NO2])^13 Fe(NO_3)_3 | 1 | 1 | [Fe(NO3)3] HIO_3 | 2 | 2 | ([HIO3])^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])^(-16) ([FeI2])^(-1) ([H2O])^7 ([NO2])^13 [Fe(NO3)3] ([HIO3])^2 = (([H2O])^7 ([NO2])^13 [Fe(NO3)3] ([HIO3])^2)/(([HNO3])^16 [FeI2])
Construct the equilibrium constant, K, expression for: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + Fe(NO_3)_3 + HIO_3 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: 16 HNO_3 + FeI_2 ⟶ 7 H_2O + 13 NO_2 + Fe(NO_3)_3 + 2 HIO_3 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 | 16 | -16 FeI_2 | 1 | -1 H_2O | 7 | 7 NO_2 | 13 | 13 Fe(NO_3)_3 | 1 | 1 HIO_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 16 | -16 | ([HNO3])^(-16) FeI_2 | 1 | -1 | ([FeI2])^(-1) H_2O | 7 | 7 | ([H2O])^7 NO_2 | 13 | 13 | ([NO2])^13 Fe(NO_3)_3 | 1 | 1 | [Fe(NO3)3] HIO_3 | 2 | 2 | ([HIO3])^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])^(-16) ([FeI2])^(-1) ([H2O])^7 ([NO2])^13 [Fe(NO3)3] ([HIO3])^2 = (([H2O])^7 ([NO2])^13 [Fe(NO3)3] ([HIO3])^2)/(([HNO3])^16 [FeI2])

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + Fe(NO_3)_3 + HIO_3 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: 16 HNO_3 + FeI_2 ⟶ 7 H_2O + 13 NO_2 + Fe(NO_3)_3 + 2 HIO_3 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 | 16 | -16 FeI_2 | 1 | -1 H_2O | 7 | 7 NO_2 | 13 | 13 Fe(NO_3)_3 | 1 | 1 HIO_3 | 2 | 2 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 | 16 | -16 | -1/16 (Δ[HNO3])/(Δt) FeI_2 | 1 | -1 | -(Δ[FeI2])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) NO_2 | 13 | 13 | 1/13 (Δ[NO2])/(Δt) Fe(NO_3)_3 | 1 | 1 | (Δ[Fe(NO3)3])/(Δt) HIO_3 | 2 | 2 | 1/2 (Δ[HIO3])/(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 = -1/16 (Δ[HNO3])/(Δt) = -(Δ[FeI2])/(Δt) = 1/7 (Δ[H2O])/(Δt) = 1/13 (Δ[NO2])/(Δt) = (Δ[Fe(NO3)3])/(Δt) = 1/2 (Δ[HIO3])/(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 + Fe(NO_3)_3 + HIO_3 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: 16 HNO_3 + FeI_2 ⟶ 7 H_2O + 13 NO_2 + Fe(NO_3)_3 + 2 HIO_3 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 | 16 | -16 FeI_2 | 1 | -1 H_2O | 7 | 7 NO_2 | 13 | 13 Fe(NO_3)_3 | 1 | 1 HIO_3 | 2 | 2 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 | 16 | -16 | -1/16 (Δ[HNO3])/(Δt) FeI_2 | 1 | -1 | -(Δ[FeI2])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) NO_2 | 13 | 13 | 1/13 (Δ[NO2])/(Δt) Fe(NO_3)_3 | 1 | 1 | (Δ[Fe(NO3)3])/(Δt) HIO_3 | 2 | 2 | 1/2 (Δ[HIO3])/(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 = -1/16 (Δ[HNO3])/(Δt) = -(Δ[FeI2])/(Δt) = 1/7 (Δ[H2O])/(Δt) = 1/13 (Δ[NO2])/(Δt) = (Δ[Fe(NO3)3])/(Δt) = 1/2 (Δ[HIO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | ferrous iodide | water | nitrogen dioxide | ferric nitrate | iodic acid formula | HNO_3 | FeI_2 | H_2O | NO_2 | Fe(NO_3)_3 | HIO_3 Hill formula | HNO_3 | FeI_2 | H_2O | NO_2 | FeN_3O_9 | HIO_3 name | nitric acid | ferrous iodide | water | nitrogen dioxide | ferric nitrate | iodic acid IUPAC name | nitric acid | diiodoiron | water | Nitrogen dioxide | iron(+3) cation trinitrate | iodic acid
| nitric acid | ferrous iodide | water | nitrogen dioxide | ferric nitrate | iodic acid formula | HNO_3 | FeI_2 | H_2O | NO_2 | Fe(NO_3)_3 | HIO_3 Hill formula | HNO_3 | FeI_2 | H_2O | NO_2 | FeN_3O_9 | HIO_3 name | nitric acid | ferrous iodide | water | nitrogen dioxide | ferric nitrate | iodic acid IUPAC name | nitric acid | diiodoiron | water | Nitrogen dioxide | iron(+3) cation trinitrate | iodic acid

Random Queries

melting point of barium molybdate vs sodium molybdate dihydrate
molar mass of tin(II) iodide
molar mass of calcium telluride vs calcium carbonate-13 C
atomic radius of lithium
molar mass of lanthanum hydride
crystal class of albite
meteorite abundance of group 14 elements
density of chromic acid vs methanesulfonic acid
IUPAC lithium fluoride
color of marshite