Search

H2O + O2 + FeOH2 = Fe(OH)3

Input interpretation

H_2O water + O_2 oxygen + FeOH2 ⟶ Fe(OH)_3 iron(III) hydroxide
H_2O water + O_2 oxygen + FeOH2 ⟶ Fe(OH)_3 iron(III) hydroxide

Balanced equation

Balance the chemical equation algebraically: H_2O + O_2 + FeOH2 ⟶ Fe(OH)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_2 + c_3 FeOH2 ⟶ c_4 Fe(OH)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and Fe: H: | 2 c_1 + 2 c_3 = 3 c_4 O: | c_1 + 2 c_2 + c_3 = 3 c_4 Fe: | c_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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3/2 c_3 = 2 c_4 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 4 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + 3 O_2 + 4 FeOH2 ⟶ 4 Fe(OH)_3
Balance the chemical equation algebraically: H_2O + O_2 + FeOH2 ⟶ Fe(OH)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_2 + c_3 FeOH2 ⟶ c_4 Fe(OH)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and Fe: H: | 2 c_1 + 2 c_3 = 3 c_4 O: | c_1 + 2 c_2 + c_3 = 3 c_4 Fe: | c_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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3/2 c_3 = 2 c_4 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 4 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + 3 O_2 + 4 FeOH2 ⟶ 4 Fe(OH)_3

Structures

+ + FeOH2 ⟶
+ + FeOH2 ⟶

Names

water + oxygen + FeOH2 ⟶ iron(III) hydroxide
water + oxygen + FeOH2 ⟶ iron(III) hydroxide

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + O_2 + FeOH2 ⟶ Fe(OH)_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: 2 H_2O + 3 O_2 + 4 FeOH2 ⟶ 4 Fe(OH)_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 H_2O | 2 | -2 O_2 | 3 | -3 FeOH2 | 4 | -4 Fe(OH)_3 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 2 | -2 | ([H2O])^(-2) O_2 | 3 | -3 | ([O2])^(-3) FeOH2 | 4 | -4 | ([FeOH2])^(-4) Fe(OH)_3 | 4 | 4 | ([Fe(OH)3])^4 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 = ([H2O])^(-2) ([O2])^(-3) ([FeOH2])^(-4) ([Fe(OH)3])^4 = ([Fe(OH)3])^4/(([H2O])^2 ([O2])^3 ([FeOH2])^4)
Construct the equilibrium constant, K, expression for: H_2O + O_2 + FeOH2 ⟶ Fe(OH)_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: 2 H_2O + 3 O_2 + 4 FeOH2 ⟶ 4 Fe(OH)_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 H_2O | 2 | -2 O_2 | 3 | -3 FeOH2 | 4 | -4 Fe(OH)_3 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 2 | -2 | ([H2O])^(-2) O_2 | 3 | -3 | ([O2])^(-3) FeOH2 | 4 | -4 | ([FeOH2])^(-4) Fe(OH)_3 | 4 | 4 | ([Fe(OH)3])^4 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 = ([H2O])^(-2) ([O2])^(-3) ([FeOH2])^(-4) ([Fe(OH)3])^4 = ([Fe(OH)3])^4/(([H2O])^2 ([O2])^3 ([FeOH2])^4)

Rate of reaction

Construct the rate of reaction expression for: H_2O + O_2 + FeOH2 ⟶ Fe(OH)_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: 2 H_2O + 3 O_2 + 4 FeOH2 ⟶ 4 Fe(OH)_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 H_2O | 2 | -2 O_2 | 3 | -3 FeOH2 | 4 | -4 Fe(OH)_3 | 4 | 4 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 H_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) O_2 | 3 | -3 | -1/3 (Δ[O2])/(Δt) FeOH2 | 4 | -4 | -1/4 (Δ[FeOH2])/(Δt) Fe(OH)_3 | 4 | 4 | 1/4 (Δ[Fe(OH)3])/(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[H2O])/(Δt) = -1/3 (Δ[O2])/(Δt) = -1/4 (Δ[FeOH2])/(Δt) = 1/4 (Δ[Fe(OH)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + O_2 + FeOH2 ⟶ Fe(OH)_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: 2 H_2O + 3 O_2 + 4 FeOH2 ⟶ 4 Fe(OH)_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 H_2O | 2 | -2 O_2 | 3 | -3 FeOH2 | 4 | -4 Fe(OH)_3 | 4 | 4 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 H_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) O_2 | 3 | -3 | -1/3 (Δ[O2])/(Δt) FeOH2 | 4 | -4 | -1/4 (Δ[FeOH2])/(Δt) Fe(OH)_3 | 4 | 4 | 1/4 (Δ[Fe(OH)3])/(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[H2O])/(Δt) = -1/3 (Δ[O2])/(Δt) = -1/4 (Δ[FeOH2])/(Δt) = 1/4 (Δ[Fe(OH)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | oxygen | FeOH2 | iron(III) hydroxide formula | H_2O | O_2 | FeOH2 | Fe(OH)_3 Hill formula | H_2O | O_2 | H2FeO | FeH_3O_3 name | water | oxygen | | iron(III) hydroxide IUPAC name | water | molecular oxygen | | ferric trihydroxide
| water | oxygen | FeOH2 | iron(III) hydroxide formula | H_2O | O_2 | FeOH2 | Fe(OH)_3 Hill formula | H_2O | O_2 | H2FeO | FeH_3O_3 name | water | oxygen | | iron(III) hydroxide IUPAC name | water | molecular oxygen | | ferric trihydroxide

Substance properties

| water | oxygen | FeOH2 | iron(III) hydroxide molar mass | 18.015 g/mol | 31.998 g/mol | 73.86 g/mol | 106.87 g/mol phase | liquid (at STP) | gas (at STP) | | melting point | 0 °C | -218 °C | | boiling point | 99.9839 °C | -183 °C | | density | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | | surface tension | 0.0728 N/m | 0.01347 N/m | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | | odor | odorless | odorless | |
| water | oxygen | FeOH2 | iron(III) hydroxide molar mass | 18.015 g/mol | 31.998 g/mol | 73.86 g/mol | 106.87 g/mol phase | liquid (at STP) | gas (at STP) | | melting point | 0 °C | -218 °C | | boiling point | 99.9839 °C | -183 °C | | density | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | | surface tension | 0.0728 N/m | 0.01347 N/m | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | | odor | odorless | odorless | |

Units

Random Queries

chemical types of cesium chlorate vs calcium oxalate hydrate
Cl2 + P4 = PCl5
ion class of fluoride anion vs iodide anion
chlorine vs gallium vs bromine vs neon
O2 + C2H6 = H2O + CO2
total number of protons of chloride anion vs arsenide anion
chemical types terbium(III) sulfate octahydrate
melting point of hexamethylmelamine
NO2 + C = CO2 + NO
molar mass of uranium(VI) cation vs erbium(III) cation