Input interpretation
![Bi(OH)3 + K2SnO2 ⟶ H_2O water + Bi bismuth + K_2O_3Sn potassium stannate](../image_source/107fa1dde66a40aa565556b71abdd547.png)
Bi(OH)3 + K2SnO2 ⟶ H_2O water + Bi bismuth + K_2O_3Sn potassium stannate
Balanced equation
![Balance the chemical equation algebraically: Bi(OH)3 + K2SnO2 ⟶ H_2O + Bi + K_2O_3Sn Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Bi(OH)3 + c_2 K2SnO2 ⟶ c_3 H_2O + c_4 Bi + c_5 K_2O_3Sn Set the number of atoms in the reactants equal to the number of atoms in the products for Bi, O, H, K and Sn: Bi: | c_1 = c_4 O: | 3 c_1 + 2 c_2 = c_3 + 3 c_5 H: | 3 c_1 = 2 c_3 K: | 2 c_2 = 2 c_5 Sn: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3/2 c_3 = 3/2 c_4 = 1 c_5 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 3 c_4 = 2 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Bi(OH)3 + 3 K2SnO2 ⟶ 3 H_2O + 2 Bi + 3 K_2O_3Sn](../image_source/908733b2e7c9f8e55394eb741e6d64f8.png)
Balance the chemical equation algebraically: Bi(OH)3 + K2SnO2 ⟶ H_2O + Bi + K_2O_3Sn Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Bi(OH)3 + c_2 K2SnO2 ⟶ c_3 H_2O + c_4 Bi + c_5 K_2O_3Sn Set the number of atoms in the reactants equal to the number of atoms in the products for Bi, O, H, K and Sn: Bi: | c_1 = c_4 O: | 3 c_1 + 2 c_2 = c_3 + 3 c_5 H: | 3 c_1 = 2 c_3 K: | 2 c_2 = 2 c_5 Sn: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3/2 c_3 = 3/2 c_4 = 1 c_5 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 3 c_4 = 2 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Bi(OH)3 + 3 K2SnO2 ⟶ 3 H_2O + 2 Bi + 3 K_2O_3Sn
Structures
![Bi(OH)3 + K2SnO2 ⟶ + +](../image_source/b2d51c06273d5e17e9c00c068c48a3df.png)
Bi(OH)3 + K2SnO2 ⟶ + +
Names
![Bi(OH)3 + K2SnO2 ⟶ water + bismuth + potassium stannate](../image_source/4784d16cd66f912803386ee6f043372d.png)
Bi(OH)3 + K2SnO2 ⟶ water + bismuth + potassium stannate
Equilibrium constant
![Construct the equilibrium constant, K, expression for: Bi(OH)3 + K2SnO2 ⟶ H_2O + Bi + K_2O_3Sn 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 Bi(OH)3 + 3 K2SnO2 ⟶ 3 H_2O + 2 Bi + 3 K_2O_3Sn 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 Bi(OH)3 | 2 | -2 K2SnO2 | 3 | -3 H_2O | 3 | 3 Bi | 2 | 2 K_2O_3Sn | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Bi(OH)3 | 2 | -2 | ([Bi(OH)3])^(-2) K2SnO2 | 3 | -3 | ([K2SnO2])^(-3) H_2O | 3 | 3 | ([H2O])^3 Bi | 2 | 2 | ([Bi])^2 K_2O_3Sn | 3 | 3 | ([K2O3Sn])^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 = ([Bi(OH)3])^(-2) ([K2SnO2])^(-3) ([H2O])^3 ([Bi])^2 ([K2O3Sn])^3 = (([H2O])^3 ([Bi])^2 ([K2O3Sn])^3)/(([Bi(OH)3])^2 ([K2SnO2])^3)](../image_source/298cc0c6f810b8347f370895c6b962c1.png)
Construct the equilibrium constant, K, expression for: Bi(OH)3 + K2SnO2 ⟶ H_2O + Bi + K_2O_3Sn 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 Bi(OH)3 + 3 K2SnO2 ⟶ 3 H_2O + 2 Bi + 3 K_2O_3Sn 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 Bi(OH)3 | 2 | -2 K2SnO2 | 3 | -3 H_2O | 3 | 3 Bi | 2 | 2 K_2O_3Sn | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Bi(OH)3 | 2 | -2 | ([Bi(OH)3])^(-2) K2SnO2 | 3 | -3 | ([K2SnO2])^(-3) H_2O | 3 | 3 | ([H2O])^3 Bi | 2 | 2 | ([Bi])^2 K_2O_3Sn | 3 | 3 | ([K2O3Sn])^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 = ([Bi(OH)3])^(-2) ([K2SnO2])^(-3) ([H2O])^3 ([Bi])^2 ([K2O3Sn])^3 = (([H2O])^3 ([Bi])^2 ([K2O3Sn])^3)/(([Bi(OH)3])^2 ([K2SnO2])^3)
Rate of reaction
![Construct the rate of reaction expression for: Bi(OH)3 + K2SnO2 ⟶ H_2O + Bi + K_2O_3Sn 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 Bi(OH)3 + 3 K2SnO2 ⟶ 3 H_2O + 2 Bi + 3 K_2O_3Sn 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 Bi(OH)3 | 2 | -2 K2SnO2 | 3 | -3 H_2O | 3 | 3 Bi | 2 | 2 K_2O_3Sn | 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 Bi(OH)3 | 2 | -2 | -1/2 (Δ[Bi(OH)3])/(Δt) K2SnO2 | 3 | -3 | -1/3 (Δ[K2SnO2])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) Bi | 2 | 2 | 1/2 (Δ[Bi])/(Δt) K_2O_3Sn | 3 | 3 | 1/3 (Δ[K2O3Sn])/(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[Bi(OH)3])/(Δt) = -1/3 (Δ[K2SnO2])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/2 (Δ[Bi])/(Δt) = 1/3 (Δ[K2O3Sn])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/6d75b4dbc9ea166eec11d7736d75d56d.png)
Construct the rate of reaction expression for: Bi(OH)3 + K2SnO2 ⟶ H_2O + Bi + K_2O_3Sn 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 Bi(OH)3 + 3 K2SnO2 ⟶ 3 H_2O + 2 Bi + 3 K_2O_3Sn 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 Bi(OH)3 | 2 | -2 K2SnO2 | 3 | -3 H_2O | 3 | 3 Bi | 2 | 2 K_2O_3Sn | 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 Bi(OH)3 | 2 | -2 | -1/2 (Δ[Bi(OH)3])/(Δt) K2SnO2 | 3 | -3 | -1/3 (Δ[K2SnO2])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) Bi | 2 | 2 | 1/2 (Δ[Bi])/(Δt) K_2O_3Sn | 3 | 3 | 1/3 (Δ[K2O3Sn])/(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[Bi(OH)3])/(Δt) = -1/3 (Δ[K2SnO2])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/2 (Δ[Bi])/(Δt) = 1/3 (Δ[K2O3Sn])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| Bi(OH)3 | K2SnO2 | water | bismuth | potassium stannate formula | Bi(OH)3 | K2SnO2 | H_2O | Bi | K_2O_3Sn Hill formula | H3BiO3 | K2O2Sn | H_2O | Bi | K_2O_3Sn name | | | water | bismuth | potassium stannate IUPAC name | | | water | bismuth | dipotassium dioxido-oxo-tin](../image_source/016e3a3cc7d1dadbe2082c0737549cab.png)
| Bi(OH)3 | K2SnO2 | water | bismuth | potassium stannate formula | Bi(OH)3 | K2SnO2 | H_2O | Bi | K_2O_3Sn Hill formula | H3BiO3 | K2O2Sn | H_2O | Bi | K_2O_3Sn name | | | water | bismuth | potassium stannate IUPAC name | | | water | bismuth | dipotassium dioxido-oxo-tin
Substance properties
![| Bi(OH)3 | K2SnO2 | water | bismuth | potassium stannate molar mass | 260.001 g/mol | 228.905 g/mol | 18.015 g/mol | 208.9804 g/mol | 244.904 g/mol phase | | | liquid (at STP) | solid (at STP) | melting point | | | 0 °C | 271 °C | boiling point | | | 99.9839 °C | 1560 °C | density | | | 1 g/cm^3 | 9.8 g/cm^3 | 3.19 g/cm^3 solubility in water | | | | insoluble | surface tension | | | 0.0728 N/m | | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | 1.19×10^-4 Pa s (at 500 °C) | odor | | | odorless | |](../image_source/997b0e3810ee3a2224631867e4131469.png)
| Bi(OH)3 | K2SnO2 | water | bismuth | potassium stannate molar mass | 260.001 g/mol | 228.905 g/mol | 18.015 g/mol | 208.9804 g/mol | 244.904 g/mol phase | | | liquid (at STP) | solid (at STP) | melting point | | | 0 °C | 271 °C | boiling point | | | 99.9839 °C | 1560 °C | density | | | 1 g/cm^3 | 9.8 g/cm^3 | 3.19 g/cm^3 solubility in water | | | | insoluble | surface tension | | | 0.0728 N/m | | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | 1.19×10^-4 Pa s (at 500 °C) | odor | | | odorless | |
Units