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KOH + KMnO4 + Sn(OH)2 = H2O + MnO2 + K2SnO3

Input interpretation

KOH potassium hydroxide + KMnO_4 potassium permanganate + Sn(OH)_2 tin(II) hydroxide ⟶ H_2O water + MnO_2 manganese dioxide + K_2O_3Sn potassium stannate
KOH potassium hydroxide + KMnO_4 potassium permanganate + Sn(OH)_2 tin(II) hydroxide ⟶ H_2O water + MnO_2 manganese dioxide + K_2O_3Sn potassium stannate

Balanced equation

Balance the chemical equation algebraically: KOH + KMnO_4 + Sn(OH)_2 ⟶ H_2O + MnO_2 + K_2O_3Sn Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 KMnO_4 + c_3 Sn(OH)_2 ⟶ c_4 H_2O + c_5 MnO_2 + c_6 K_2O_3Sn Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Mn and Sn: H: | c_1 + 2 c_3 = 2 c_4 K: | c_1 + c_2 = 2 c_6 O: | c_1 + 4 c_2 + 2 c_3 = c_4 + 2 c_5 + 3 c_6 Mn: | c_2 = c_5 Sn: | 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 = 3/2 c_4 = 5/2 c_5 = 1 c_6 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 2 c_3 = 3 c_4 = 5 c_5 = 2 c_6 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 KOH + 2 KMnO_4 + 3 Sn(OH)_2 ⟶ 5 H_2O + 2 MnO_2 + 3 K_2O_3Sn
Balance the chemical equation algebraically: KOH + KMnO_4 + Sn(OH)_2 ⟶ H_2O + MnO_2 + K_2O_3Sn Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 KMnO_4 + c_3 Sn(OH)_2 ⟶ c_4 H_2O + c_5 MnO_2 + c_6 K_2O_3Sn Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Mn and Sn: H: | c_1 + 2 c_3 = 2 c_4 K: | c_1 + c_2 = 2 c_6 O: | c_1 + 4 c_2 + 2 c_3 = c_4 + 2 c_5 + 3 c_6 Mn: | c_2 = c_5 Sn: | 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 = 3/2 c_4 = 5/2 c_5 = 1 c_6 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 2 c_3 = 3 c_4 = 5 c_5 = 2 c_6 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 KOH + 2 KMnO_4 + 3 Sn(OH)_2 ⟶ 5 H_2O + 2 MnO_2 + 3 K_2O_3Sn

Structures

+ + ⟶ + +
+ + ⟶ + +

Names

potassium hydroxide + potassium permanganate + tin(II) hydroxide ⟶ water + manganese dioxide + potassium stannate
potassium hydroxide + potassium permanganate + tin(II) hydroxide ⟶ water + manganese dioxide + potassium stannate

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + KMnO_4 + Sn(OH)_2 ⟶ H_2O + MnO_2 + 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: 4 KOH + 2 KMnO_4 + 3 Sn(OH)_2 ⟶ 5 H_2O + 2 MnO_2 + 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 KOH | 4 | -4 KMnO_4 | 2 | -2 Sn(OH)_2 | 3 | -3 H_2O | 5 | 5 MnO_2 | 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 KOH | 4 | -4 | ([KOH])^(-4) KMnO_4 | 2 | -2 | ([KMnO4])^(-2) Sn(OH)_2 | 3 | -3 | ([Sn(OH)2])^(-3) H_2O | 5 | 5 | ([H2O])^5 MnO_2 | 2 | 2 | ([MnO2])^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 = ([KOH])^(-4) ([KMnO4])^(-2) ([Sn(OH)2])^(-3) ([H2O])^5 ([MnO2])^2 ([K2O3Sn])^3 = (([H2O])^5 ([MnO2])^2 ([K2O3Sn])^3)/(([KOH])^4 ([KMnO4])^2 ([Sn(OH)2])^3)
Construct the equilibrium constant, K, expression for: KOH + KMnO_4 + Sn(OH)_2 ⟶ H_2O + MnO_2 + 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: 4 KOH + 2 KMnO_4 + 3 Sn(OH)_2 ⟶ 5 H_2O + 2 MnO_2 + 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 KOH | 4 | -4 KMnO_4 | 2 | -2 Sn(OH)_2 | 3 | -3 H_2O | 5 | 5 MnO_2 | 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 KOH | 4 | -4 | ([KOH])^(-4) KMnO_4 | 2 | -2 | ([KMnO4])^(-2) Sn(OH)_2 | 3 | -3 | ([Sn(OH)2])^(-3) H_2O | 5 | 5 | ([H2O])^5 MnO_2 | 2 | 2 | ([MnO2])^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 = ([KOH])^(-4) ([KMnO4])^(-2) ([Sn(OH)2])^(-3) ([H2O])^5 ([MnO2])^2 ([K2O3Sn])^3 = (([H2O])^5 ([MnO2])^2 ([K2O3Sn])^3)/(([KOH])^4 ([KMnO4])^2 ([Sn(OH)2])^3)

Rate of reaction

Construct the rate of reaction expression for: KOH + KMnO_4 + Sn(OH)_2 ⟶ H_2O + MnO_2 + 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: 4 KOH + 2 KMnO_4 + 3 Sn(OH)_2 ⟶ 5 H_2O + 2 MnO_2 + 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 KOH | 4 | -4 KMnO_4 | 2 | -2 Sn(OH)_2 | 3 | -3 H_2O | 5 | 5 MnO_2 | 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 KOH | 4 | -4 | -1/4 (Δ[KOH])/(Δt) KMnO_4 | 2 | -2 | -1/2 (Δ[KMnO4])/(Δt) Sn(OH)_2 | 3 | -3 | -1/3 (Δ[Sn(OH)2])/(Δt) H_2O | 5 | 5 | 1/5 (Δ[H2O])/(Δt) MnO_2 | 2 | 2 | 1/2 (Δ[MnO2])/(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 = -1/4 (Δ[KOH])/(Δt) = -1/2 (Δ[KMnO4])/(Δt) = -1/3 (Δ[Sn(OH)2])/(Δt) = 1/5 (Δ[H2O])/(Δt) = 1/2 (Δ[MnO2])/(Δt) = 1/3 (Δ[K2O3Sn])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + KMnO_4 + Sn(OH)_2 ⟶ H_2O + MnO_2 + 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: 4 KOH + 2 KMnO_4 + 3 Sn(OH)_2 ⟶ 5 H_2O + 2 MnO_2 + 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 KOH | 4 | -4 KMnO_4 | 2 | -2 Sn(OH)_2 | 3 | -3 H_2O | 5 | 5 MnO_2 | 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 KOH | 4 | -4 | -1/4 (Δ[KOH])/(Δt) KMnO_4 | 2 | -2 | -1/2 (Δ[KMnO4])/(Δt) Sn(OH)_2 | 3 | -3 | -1/3 (Δ[Sn(OH)2])/(Δt) H_2O | 5 | 5 | 1/5 (Δ[H2O])/(Δt) MnO_2 | 2 | 2 | 1/2 (Δ[MnO2])/(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 = -1/4 (Δ[KOH])/(Δt) = -1/2 (Δ[KMnO4])/(Δt) = -1/3 (Δ[Sn(OH)2])/(Δt) = 1/5 (Δ[H2O])/(Δt) = 1/2 (Δ[MnO2])/(Δt) = 1/3 (Δ[K2O3Sn])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium hydroxide | potassium permanganate | tin(II) hydroxide | water | manganese dioxide | potassium stannate formula | KOH | KMnO_4 | Sn(OH)_2 | H_2O | MnO_2 | K_2O_3Sn Hill formula | HKO | KMnO_4 | H_2O_2Sn | H_2O | MnO_2 | K_2O_3Sn name | potassium hydroxide | potassium permanganate | tin(II) hydroxide | water | manganese dioxide | potassium stannate IUPAC name | potassium hydroxide | potassium permanganate | stannous dihydroxide | water | dioxomanganese | dipotassium dioxido-oxo-tin
| potassium hydroxide | potassium permanganate | tin(II) hydroxide | water | manganese dioxide | potassium stannate formula | KOH | KMnO_4 | Sn(OH)_2 | H_2O | MnO_2 | K_2O_3Sn Hill formula | HKO | KMnO_4 | H_2O_2Sn | H_2O | MnO_2 | K_2O_3Sn name | potassium hydroxide | potassium permanganate | tin(II) hydroxide | water | manganese dioxide | potassium stannate IUPAC name | potassium hydroxide | potassium permanganate | stannous dihydroxide | water | dioxomanganese | dipotassium dioxido-oxo-tin

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