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HNO3 + Mn(NO3)2 + Na3BiO4 = H2O + NaNO3 + HMnO4 + Bi(NO3)3

Input interpretation

HNO_3 nitric acid + Mn(NO_3)_2 manganese(II) nitrate + Na3BiO4 ⟶ H_2O water + NaNO_3 sodium nitrate + HMnO4 + Bi(NO3)3
HNO_3 nitric acid + Mn(NO_3)_2 manganese(II) nitrate + Na3BiO4 ⟶ H_2O water + NaNO_3 sodium nitrate + HMnO4 + Bi(NO3)3

Balanced equation

Balance the chemical equation algebraically: HNO_3 + Mn(NO_3)_2 + Na3BiO4 ⟶ H_2O + NaNO_3 + HMnO4 + Bi(NO3)3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Mn(NO_3)_2 + c_3 Na3BiO4 ⟶ c_4 H_2O + c_5 NaNO_3 + c_6 HMnO4 + c_7 Bi(NO3)3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Mn, Na and Bi: H: | c_1 = 2 c_4 + c_6 N: | c_1 + 2 c_2 = c_5 + 3 c_7 O: | 3 c_1 + 6 c_2 + 4 c_3 = c_4 + 3 c_5 + 4 c_6 + 9 c_7 Mn: | c_2 = c_6 Na: | 3 c_3 = c_5 Bi: | c_3 = c_7 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 = 13 c_2 = 1 c_3 = 5/2 c_4 = 6 c_5 = 15/2 c_6 = 1 c_7 = 5/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 26 c_2 = 2 c_3 = 5 c_4 = 12 c_5 = 15 c_6 = 2 c_7 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 26 HNO_3 + 2 Mn(NO_3)_2 + 5 Na3BiO4 ⟶ 12 H_2O + 15 NaNO_3 + 2 HMnO4 + 5 Bi(NO3)3
Balance the chemical equation algebraically: HNO_3 + Mn(NO_3)_2 + Na3BiO4 ⟶ H_2O + NaNO_3 + HMnO4 + Bi(NO3)3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Mn(NO_3)_2 + c_3 Na3BiO4 ⟶ c_4 H_2O + c_5 NaNO_3 + c_6 HMnO4 + c_7 Bi(NO3)3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Mn, Na and Bi: H: | c_1 = 2 c_4 + c_6 N: | c_1 + 2 c_2 = c_5 + 3 c_7 O: | 3 c_1 + 6 c_2 + 4 c_3 = c_4 + 3 c_5 + 4 c_6 + 9 c_7 Mn: | c_2 = c_6 Na: | 3 c_3 = c_5 Bi: | c_3 = c_7 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 = 13 c_2 = 1 c_3 = 5/2 c_4 = 6 c_5 = 15/2 c_6 = 1 c_7 = 5/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 26 c_2 = 2 c_3 = 5 c_4 = 12 c_5 = 15 c_6 = 2 c_7 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 26 HNO_3 + 2 Mn(NO_3)_2 + 5 Na3BiO4 ⟶ 12 H_2O + 15 NaNO_3 + 2 HMnO4 + 5 Bi(NO3)3

Structures

+ + Na3BiO4 ⟶ + + HMnO4 + Bi(NO3)3
+ + Na3BiO4 ⟶ + + HMnO4 + Bi(NO3)3

Names

nitric acid + manganese(II) nitrate + Na3BiO4 ⟶ water + sodium nitrate + HMnO4 + Bi(NO3)3
nitric acid + manganese(II) nitrate + Na3BiO4 ⟶ water + sodium nitrate + HMnO4 + Bi(NO3)3

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + Mn(NO_3)_2 + Na3BiO4 ⟶ H_2O + NaNO_3 + HMnO4 + Bi(NO3)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: 26 HNO_3 + 2 Mn(NO_3)_2 + 5 Na3BiO4 ⟶ 12 H_2O + 15 NaNO_3 + 2 HMnO4 + 5 Bi(NO3)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 | 26 | -26 Mn(NO_3)_2 | 2 | -2 Na3BiO4 | 5 | -5 H_2O | 12 | 12 NaNO_3 | 15 | 15 HMnO4 | 2 | 2 Bi(NO3)3 | 5 | 5 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 26 | -26 | ([HNO3])^(-26) Mn(NO_3)_2 | 2 | -2 | ([Mn(NO3)2])^(-2) Na3BiO4 | 5 | -5 | ([Na3BiO4])^(-5) H_2O | 12 | 12 | ([H2O])^12 NaNO_3 | 15 | 15 | ([NaNO3])^15 HMnO4 | 2 | 2 | ([HMnO4])^2 Bi(NO3)3 | 5 | 5 | ([Bi(NO3)3])^5 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])^(-26) ([Mn(NO3)2])^(-2) ([Na3BiO4])^(-5) ([H2O])^12 ([NaNO3])^15 ([HMnO4])^2 ([Bi(NO3)3])^5 = (([H2O])^12 ([NaNO3])^15 ([HMnO4])^2 ([Bi(NO3)3])^5)/(([HNO3])^26 ([Mn(NO3)2])^2 ([Na3BiO4])^5)
Construct the equilibrium constant, K, expression for: HNO_3 + Mn(NO_3)_2 + Na3BiO4 ⟶ H_2O + NaNO_3 + HMnO4 + Bi(NO3)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: 26 HNO_3 + 2 Mn(NO_3)_2 + 5 Na3BiO4 ⟶ 12 H_2O + 15 NaNO_3 + 2 HMnO4 + 5 Bi(NO3)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 | 26 | -26 Mn(NO_3)_2 | 2 | -2 Na3BiO4 | 5 | -5 H_2O | 12 | 12 NaNO_3 | 15 | 15 HMnO4 | 2 | 2 Bi(NO3)3 | 5 | 5 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 26 | -26 | ([HNO3])^(-26) Mn(NO_3)_2 | 2 | -2 | ([Mn(NO3)2])^(-2) Na3BiO4 | 5 | -5 | ([Na3BiO4])^(-5) H_2O | 12 | 12 | ([H2O])^12 NaNO_3 | 15 | 15 | ([NaNO3])^15 HMnO4 | 2 | 2 | ([HMnO4])^2 Bi(NO3)3 | 5 | 5 | ([Bi(NO3)3])^5 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])^(-26) ([Mn(NO3)2])^(-2) ([Na3BiO4])^(-5) ([H2O])^12 ([NaNO3])^15 ([HMnO4])^2 ([Bi(NO3)3])^5 = (([H2O])^12 ([NaNO3])^15 ([HMnO4])^2 ([Bi(NO3)3])^5)/(([HNO3])^26 ([Mn(NO3)2])^2 ([Na3BiO4])^5)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + Mn(NO_3)_2 + Na3BiO4 ⟶ H_2O + NaNO_3 + HMnO4 + Bi(NO3)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: 26 HNO_3 + 2 Mn(NO_3)_2 + 5 Na3BiO4 ⟶ 12 H_2O + 15 NaNO_3 + 2 HMnO4 + 5 Bi(NO3)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 | 26 | -26 Mn(NO_3)_2 | 2 | -2 Na3BiO4 | 5 | -5 H_2O | 12 | 12 NaNO_3 | 15 | 15 HMnO4 | 2 | 2 Bi(NO3)3 | 5 | 5 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 | 26 | -26 | -1/26 (Δ[HNO3])/(Δt) Mn(NO_3)_2 | 2 | -2 | -1/2 (Δ[Mn(NO3)2])/(Δt) Na3BiO4 | 5 | -5 | -1/5 (Δ[Na3BiO4])/(Δt) H_2O | 12 | 12 | 1/12 (Δ[H2O])/(Δt) NaNO_3 | 15 | 15 | 1/15 (Δ[NaNO3])/(Δt) HMnO4 | 2 | 2 | 1/2 (Δ[HMnO4])/(Δt) Bi(NO3)3 | 5 | 5 | 1/5 (Δ[Bi(NO3)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 = -1/26 (Δ[HNO3])/(Δt) = -1/2 (Δ[Mn(NO3)2])/(Δt) = -1/5 (Δ[Na3BiO4])/(Δt) = 1/12 (Δ[H2O])/(Δt) = 1/15 (Δ[NaNO3])/(Δt) = 1/2 (Δ[HMnO4])/(Δt) = 1/5 (Δ[Bi(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + Mn(NO_3)_2 + Na3BiO4 ⟶ H_2O + NaNO_3 + HMnO4 + Bi(NO3)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: 26 HNO_3 + 2 Mn(NO_3)_2 + 5 Na3BiO4 ⟶ 12 H_2O + 15 NaNO_3 + 2 HMnO4 + 5 Bi(NO3)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 | 26 | -26 Mn(NO_3)_2 | 2 | -2 Na3BiO4 | 5 | -5 H_2O | 12 | 12 NaNO_3 | 15 | 15 HMnO4 | 2 | 2 Bi(NO3)3 | 5 | 5 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 | 26 | -26 | -1/26 (Δ[HNO3])/(Δt) Mn(NO_3)_2 | 2 | -2 | -1/2 (Δ[Mn(NO3)2])/(Δt) Na3BiO4 | 5 | -5 | -1/5 (Δ[Na3BiO4])/(Δt) H_2O | 12 | 12 | 1/12 (Δ[H2O])/(Δt) NaNO_3 | 15 | 15 | 1/15 (Δ[NaNO3])/(Δt) HMnO4 | 2 | 2 | 1/2 (Δ[HMnO4])/(Δt) Bi(NO3)3 | 5 | 5 | 1/5 (Δ[Bi(NO3)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 = -1/26 (Δ[HNO3])/(Δt) = -1/2 (Δ[Mn(NO3)2])/(Δt) = -1/5 (Δ[Na3BiO4])/(Δt) = 1/12 (Δ[H2O])/(Δt) = 1/15 (Δ[NaNO3])/(Δt) = 1/2 (Δ[HMnO4])/(Δt) = 1/5 (Δ[Bi(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | manganese(II) nitrate | Na3BiO4 | water | sodium nitrate | HMnO4 | Bi(NO3)3 formula | HNO_3 | Mn(NO_3)_2 | Na3BiO4 | H_2O | NaNO_3 | HMnO4 | Bi(NO3)3 Hill formula | HNO_3 | MnN_2O_6 | BiNa3O4 | H_2O | NNaO_3 | HMnO4 | BiN3O9 name | nitric acid | manganese(II) nitrate | | water | sodium nitrate | | IUPAC name | nitric acid | manganese(2+) dinitrate | | water | sodium nitrate | |
| nitric acid | manganese(II) nitrate | Na3BiO4 | water | sodium nitrate | HMnO4 | Bi(NO3)3 formula | HNO_3 | Mn(NO_3)_2 | Na3BiO4 | H_2O | NaNO_3 | HMnO4 | Bi(NO3)3 Hill formula | HNO_3 | MnN_2O_6 | BiNa3O4 | H_2O | NNaO_3 | HMnO4 | BiN3O9 name | nitric acid | manganese(II) nitrate | | water | sodium nitrate | | IUPAC name | nitric acid | manganese(2+) dinitrate | | water | sodium nitrate | |

Substance properties

| nitric acid | manganese(II) nitrate | Na3BiO4 | water | sodium nitrate | HMnO4 | Bi(NO3)3 molar mass | 63.012 g/mol | 178.95 g/mol | 341.946 g/mol | 18.015 g/mol | 84.994 g/mol | 119.94 g/mol | 394.99 g/mol phase | liquid (at STP) | | | liquid (at STP) | solid (at STP) | | melting point | -41.6 °C | | | 0 °C | 306 °C | | boiling point | 83 °C | | | 99.9839 °C | | | density | 1.5129 g/cm^3 | 1.536 g/cm^3 | | 1 g/cm^3 | 2.26 g/cm^3 | | solubility in water | miscible | | | | soluble | | surface tension | | | | 0.0728 N/m | | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | | 8.9×10^-4 Pa s (at 25 °C) | 0.003 Pa s (at 250 °C) | | odor | | | | odorless | | |
| nitric acid | manganese(II) nitrate | Na3BiO4 | water | sodium nitrate | HMnO4 | Bi(NO3)3 molar mass | 63.012 g/mol | 178.95 g/mol | 341.946 g/mol | 18.015 g/mol | 84.994 g/mol | 119.94 g/mol | 394.99 g/mol phase | liquid (at STP) | | | liquid (at STP) | solid (at STP) | | melting point | -41.6 °C | | | 0 °C | 306 °C | | boiling point | 83 °C | | | 99.9839 °C | | | density | 1.5129 g/cm^3 | 1.536 g/cm^3 | | 1 g/cm^3 | 2.26 g/cm^3 | | solubility in water | miscible | | | | soluble | | surface tension | | | | 0.0728 N/m | | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | | 8.9×10^-4 Pa s (at 25 °C) | 0.003 Pa s (at 250 °C) | | odor | | | | odorless | | |

Units

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