Search

KOH + I2 + K3AsO3 = H2O + KI + K3AsO4

Input interpretation

KOH potassium hydroxide + I_2 iodine + K3AsO3 ⟶ H_2O water + KI potassium iodide + K3AsO4
KOH potassium hydroxide + I_2 iodine + K3AsO3 ⟶ H_2O water + KI potassium iodide + K3AsO4

Balanced equation

Balance the chemical equation algebraically: KOH + I_2 + K3AsO3 ⟶ H_2O + KI + K3AsO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 I_2 + c_3 K3AsO3 ⟶ c_4 H_2O + c_5 KI + c_6 K3AsO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, I and As: H: | c_1 = 2 c_4 K: | c_1 + 3 c_3 = c_5 + 3 c_6 O: | c_1 + 3 c_3 = c_4 + 4 c_6 I: | 2 c_2 = c_5 As: | c_3 = 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 = 2 c_2 = 1 c_3 = 1 c_4 = 1 c_5 = 2 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + I_2 + K3AsO3 ⟶ H_2O + 2 KI + K3AsO4
Balance the chemical equation algebraically: KOH + I_2 + K3AsO3 ⟶ H_2O + KI + K3AsO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 I_2 + c_3 K3AsO3 ⟶ c_4 H_2O + c_5 KI + c_6 K3AsO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, I and As: H: | c_1 = 2 c_4 K: | c_1 + 3 c_3 = c_5 + 3 c_6 O: | c_1 + 3 c_3 = c_4 + 4 c_6 I: | 2 c_2 = c_5 As: | c_3 = 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 = 2 c_2 = 1 c_3 = 1 c_4 = 1 c_5 = 2 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + I_2 + K3AsO3 ⟶ H_2O + 2 KI + K3AsO4

Structures

+ + K3AsO3 ⟶ + + K3AsO4
+ + K3AsO3 ⟶ + + K3AsO4

Names

potassium hydroxide + iodine + K3AsO3 ⟶ water + potassium iodide + K3AsO4
potassium hydroxide + iodine + K3AsO3 ⟶ water + potassium iodide + K3AsO4

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + I_2 + K3AsO3 ⟶ H_2O + KI + K3AsO4 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 KOH + I_2 + K3AsO3 ⟶ H_2O + 2 KI + K3AsO4 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 KOH | 2 | -2 I_2 | 1 | -1 K3AsO3 | 1 | -1 H_2O | 1 | 1 KI | 2 | 2 K3AsO4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 2 | -2 | ([KOH])^(-2) I_2 | 1 | -1 | ([I2])^(-1) K3AsO3 | 1 | -1 | ([K3AsO3])^(-1) H_2O | 1 | 1 | [H2O] KI | 2 | 2 | ([KI])^2 K3AsO4 | 1 | 1 | [K3AsO4] 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 = ([KOH])^(-2) ([I2])^(-1) ([K3AsO3])^(-1) [H2O] ([KI])^2 [K3AsO4] = ([H2O] ([KI])^2 [K3AsO4])/(([KOH])^2 [I2] [K3AsO3])
Construct the equilibrium constant, K, expression for: KOH + I_2 + K3AsO3 ⟶ H_2O + KI + K3AsO4 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 KOH + I_2 + K3AsO3 ⟶ H_2O + 2 KI + K3AsO4 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 KOH | 2 | -2 I_2 | 1 | -1 K3AsO3 | 1 | -1 H_2O | 1 | 1 KI | 2 | 2 K3AsO4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 2 | -2 | ([KOH])^(-2) I_2 | 1 | -1 | ([I2])^(-1) K3AsO3 | 1 | -1 | ([K3AsO3])^(-1) H_2O | 1 | 1 | [H2O] KI | 2 | 2 | ([KI])^2 K3AsO4 | 1 | 1 | [K3AsO4] 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 = ([KOH])^(-2) ([I2])^(-1) ([K3AsO3])^(-1) [H2O] ([KI])^2 [K3AsO4] = ([H2O] ([KI])^2 [K3AsO4])/(([KOH])^2 [I2] [K3AsO3])

Rate of reaction

Construct the rate of reaction expression for: KOH + I_2 + K3AsO3 ⟶ H_2O + KI + K3AsO4 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 KOH + I_2 + K3AsO3 ⟶ H_2O + 2 KI + K3AsO4 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 KOH | 2 | -2 I_2 | 1 | -1 K3AsO3 | 1 | -1 H_2O | 1 | 1 KI | 2 | 2 K3AsO4 | 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 KOH | 2 | -2 | -1/2 (Δ[KOH])/(Δt) I_2 | 1 | -1 | -(Δ[I2])/(Δt) K3AsO3 | 1 | -1 | -(Δ[K3AsO3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) KI | 2 | 2 | 1/2 (Δ[KI])/(Δt) K3AsO4 | 1 | 1 | (Δ[K3AsO4])/(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[KOH])/(Δt) = -(Δ[I2])/(Δt) = -(Δ[K3AsO3])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[KI])/(Δt) = (Δ[K3AsO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + I_2 + K3AsO3 ⟶ H_2O + KI + K3AsO4 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 KOH + I_2 + K3AsO3 ⟶ H_2O + 2 KI + K3AsO4 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 KOH | 2 | -2 I_2 | 1 | -1 K3AsO3 | 1 | -1 H_2O | 1 | 1 KI | 2 | 2 K3AsO4 | 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 KOH | 2 | -2 | -1/2 (Δ[KOH])/(Δt) I_2 | 1 | -1 | -(Δ[I2])/(Δt) K3AsO3 | 1 | -1 | -(Δ[K3AsO3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) KI | 2 | 2 | 1/2 (Δ[KI])/(Δt) K3AsO4 | 1 | 1 | (Δ[K3AsO4])/(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[KOH])/(Δt) = -(Δ[I2])/(Δt) = -(Δ[K3AsO3])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[KI])/(Δt) = (Δ[K3AsO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium hydroxide | iodine | K3AsO3 | water | potassium iodide | K3AsO4 formula | KOH | I_2 | K3AsO3 | H_2O | KI | K3AsO4 Hill formula | HKO | I_2 | AsK3O3 | H_2O | IK | AsK3O4 name | potassium hydroxide | iodine | | water | potassium iodide | IUPAC name | potassium hydroxide | molecular iodine | | water | potassium iodide |
| potassium hydroxide | iodine | K3AsO3 | water | potassium iodide | K3AsO4 formula | KOH | I_2 | K3AsO3 | H_2O | KI | K3AsO4 Hill formula | HKO | I_2 | AsK3O3 | H_2O | IK | AsK3O4 name | potassium hydroxide | iodine | | water | potassium iodide | IUPAC name | potassium hydroxide | molecular iodine | | water | potassium iodide |

Substance properties

| potassium hydroxide | iodine | K3AsO3 | water | potassium iodide | K3AsO4 molar mass | 56.105 g/mol | 253.80894 g/mol | 240.213 g/mol | 18.015 g/mol | 166.0028 g/mol | 256.212 g/mol phase | solid (at STP) | solid (at STP) | | liquid (at STP) | solid (at STP) | melting point | 406 °C | 113 °C | | 0 °C | 681 °C | boiling point | 1327 °C | 184 °C | | 99.9839 °C | 1330 °C | density | 2.044 g/cm^3 | 4.94 g/cm^3 | | 1 g/cm^3 | 3.123 g/cm^3 | solubility in water | soluble | | | | | surface tension | | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | 0.00227 Pa s (at 116 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 0.0010227 Pa s (at 732.9 °C) | odor | | | | odorless | |
| potassium hydroxide | iodine | K3AsO3 | water | potassium iodide | K3AsO4 molar mass | 56.105 g/mol | 253.80894 g/mol | 240.213 g/mol | 18.015 g/mol | 166.0028 g/mol | 256.212 g/mol phase | solid (at STP) | solid (at STP) | | liquid (at STP) | solid (at STP) | melting point | 406 °C | 113 °C | | 0 °C | 681 °C | boiling point | 1327 °C | 184 °C | | 99.9839 °C | 1330 °C | density | 2.044 g/cm^3 | 4.94 g/cm^3 | | 1 g/cm^3 | 3.123 g/cm^3 | solubility in water | soluble | | | | | surface tension | | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | 0.00227 Pa s (at 116 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 0.0010227 Pa s (at 732.9 °C) | odor | | | | odorless | |

Units

Random Queries

name of manganese vs barium hexafluorogermanate
chemical types of betaine
1,2-propanediol dibenzoate
1,1,1-trichloroethane vs 1,1,1-trifluoroethane
molar mass of ethylene glycol vs methanol
molar mass of lithium chloride vs potassium chloride
melting point of calcium chloride
MDL number of phosphorus pentabromide vs 2-[4-(bromomethyl)phenyl]propanoic acid
mass magnetic susceptibility of lanthanum vs ytterbium
melting point of sodium bromate