Search

HNO3 + Na = H2O + NH3NO3 + NaNO4

Input interpretation

HNO_3 nitric acid + Na sodium ⟶ H_2O water + NH3NO3 + NaNO4
HNO_3 nitric acid + Na sodium ⟶ H_2O water + NH3NO3 + NaNO4

Balanced equation

Balance the chemical equation algebraically: HNO_3 + Na ⟶ H_2O + NH3NO3 + NaNO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Na ⟶ c_3 H_2O + c_4 NH3NO3 + c_5 NaNO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Na: H: | c_1 = 2 c_3 + 3 c_4 N: | c_1 = 2 c_4 + c_5 O: | 3 c_1 = c_3 + 3 c_4 + 4 c_5 Na: | 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 13/2 c_2 = 7/2 c_3 = 1 c_4 = 3/2 c_5 = 7/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 13 c_2 = 7 c_3 = 2 c_4 = 3 c_5 = 7 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 13 HNO_3 + 7 Na ⟶ 2 H_2O + 3 NH3NO3 + 7 NaNO4
Balance the chemical equation algebraically: HNO_3 + Na ⟶ H_2O + NH3NO3 + NaNO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Na ⟶ c_3 H_2O + c_4 NH3NO3 + c_5 NaNO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Na: H: | c_1 = 2 c_3 + 3 c_4 N: | c_1 = 2 c_4 + c_5 O: | 3 c_1 = c_3 + 3 c_4 + 4 c_5 Na: | 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 13/2 c_2 = 7/2 c_3 = 1 c_4 = 3/2 c_5 = 7/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 13 c_2 = 7 c_3 = 2 c_4 = 3 c_5 = 7 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 13 HNO_3 + 7 Na ⟶ 2 H_2O + 3 NH3NO3 + 7 NaNO4

Structures

 + ⟶ + NH3NO3 + NaNO4
+ ⟶ + NH3NO3 + NaNO4

Names

nitric acid + sodium ⟶ water + NH3NO3 + NaNO4
nitric acid + sodium ⟶ water + NH3NO3 + NaNO4

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + Na ⟶ H_2O + NH3NO3 + NaNO4 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: 13 HNO_3 + 7 Na ⟶ 2 H_2O + 3 NH3NO3 + 7 NaNO4 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 | 13 | -13 Na | 7 | -7 H_2O | 2 | 2 NH3NO3 | 3 | 3 NaNO4 | 7 | 7 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 13 | -13 | ([HNO3])^(-13) Na | 7 | -7 | ([Na])^(-7) H_2O | 2 | 2 | ([H2O])^2 NH3NO3 | 3 | 3 | ([NH3NO3])^3 NaNO4 | 7 | 7 | ([NaNO4])^7 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])^(-13) ([Na])^(-7) ([H2O])^2 ([NH3NO3])^3 ([NaNO4])^7 = (([H2O])^2 ([NH3NO3])^3 ([NaNO4])^7)/(([HNO3])^13 ([Na])^7)
Construct the equilibrium constant, K, expression for: HNO_3 + Na ⟶ H_2O + NH3NO3 + NaNO4 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: 13 HNO_3 + 7 Na ⟶ 2 H_2O + 3 NH3NO3 + 7 NaNO4 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 | 13 | -13 Na | 7 | -7 H_2O | 2 | 2 NH3NO3 | 3 | 3 NaNO4 | 7 | 7 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 13 | -13 | ([HNO3])^(-13) Na | 7 | -7 | ([Na])^(-7) H_2O | 2 | 2 | ([H2O])^2 NH3NO3 | 3 | 3 | ([NH3NO3])^3 NaNO4 | 7 | 7 | ([NaNO4])^7 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])^(-13) ([Na])^(-7) ([H2O])^2 ([NH3NO3])^3 ([NaNO4])^7 = (([H2O])^2 ([NH3NO3])^3 ([NaNO4])^7)/(([HNO3])^13 ([Na])^7)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + Na ⟶ H_2O + NH3NO3 + NaNO4 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: 13 HNO_3 + 7 Na ⟶ 2 H_2O + 3 NH3NO3 + 7 NaNO4 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 | 13 | -13 Na | 7 | -7 H_2O | 2 | 2 NH3NO3 | 3 | 3 NaNO4 | 7 | 7 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 | 13 | -13 | -1/13 (Δ[HNO3])/(Δt) Na | 7 | -7 | -1/7 (Δ[Na])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) NH3NO3 | 3 | 3 | 1/3 (Δ[NH3NO3])/(Δt) NaNO4 | 7 | 7 | 1/7 (Δ[NaNO4])/(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 = -1/13 (Δ[HNO3])/(Δt) = -1/7 (Δ[Na])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/3 (Δ[NH3NO3])/(Δt) = 1/7 (Δ[NaNO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + Na ⟶ H_2O + NH3NO3 + NaNO4 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: 13 HNO_3 + 7 Na ⟶ 2 H_2O + 3 NH3NO3 + 7 NaNO4 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 | 13 | -13 Na | 7 | -7 H_2O | 2 | 2 NH3NO3 | 3 | 3 NaNO4 | 7 | 7 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 | 13 | -13 | -1/13 (Δ[HNO3])/(Δt) Na | 7 | -7 | -1/7 (Δ[Na])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) NH3NO3 | 3 | 3 | 1/3 (Δ[NH3NO3])/(Δt) NaNO4 | 7 | 7 | 1/7 (Δ[NaNO4])/(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 = -1/13 (Δ[HNO3])/(Δt) = -1/7 (Δ[Na])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/3 (Δ[NH3NO3])/(Δt) = 1/7 (Δ[NaNO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | nitric acid | sodium | water | NH3NO3 | NaNO4 formula | HNO_3 | Na | H_2O | NH3NO3 | NaNO4 Hill formula | HNO_3 | Na | H_2O | H3N2O3 | NNaO4 name | nitric acid | sodium | water | |
| nitric acid | sodium | water | NH3NO3 | NaNO4 formula | HNO_3 | Na | H_2O | NH3NO3 | NaNO4 Hill formula | HNO_3 | Na | H_2O | H3N2O3 | NNaO4 name | nitric acid | sodium | water | |

Substance properties

 | nitric acid | sodium | water | NH3NO3 | NaNO4 molar mass | 63.012 g/mol | 22.98976928 g/mol | 18.015 g/mol | 79.04 g/mol | 100.99 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | |  melting point | -41.6 °C | 97.8 °C | 0 °C | |  boiling point | 83 °C | 883 °C | 99.9839 °C | |  density | 1.5129 g/cm^3 | 0.968 g/cm^3 | 1 g/cm^3 | |  solubility in water | miscible | decomposes | | |  surface tension | | | 0.0728 N/m | |  dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 1.413×10^-5 Pa s (at 527 °C) | 8.9×10^-4 Pa s (at 25 °C) | |  odor | | | odorless | |
| nitric acid | sodium | water | NH3NO3 | NaNO4 molar mass | 63.012 g/mol | 22.98976928 g/mol | 18.015 g/mol | 79.04 g/mol | 100.99 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | | melting point | -41.6 °C | 97.8 °C | 0 °C | | boiling point | 83 °C | 883 °C | 99.9839 °C | | density | 1.5129 g/cm^3 | 0.968 g/cm^3 | 1 g/cm^3 | | solubility in water | miscible | decomposes | | | surface tension | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 1.413×10^-5 Pa s (at 527 °C) | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |

Units