Search

HNO3 + Sb2S3 = H2O + H2SO4 + NO + HSbO4

Input interpretation

HNO_3 nitric acid + Sb_2S_3 antimony(III) sulfide ⟶ H_2O water + H_2SO_4 sulfuric acid + NO nitric oxide + HSbO4
HNO_3 nitric acid + Sb_2S_3 antimony(III) sulfide ⟶ H_2O water + H_2SO_4 sulfuric acid + NO nitric oxide + HSbO4

Balanced equation

Balance the chemical equation algebraically: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Sb_2S_3 ⟶ c_3 H_2O + c_4 H_2SO_4 + c_5 NO + c_6 HSbO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, S and Sb: H: | c_1 = 2 c_3 + 2 c_4 + c_6 N: | c_1 = c_5 O: | 3 c_1 = c_3 + 4 c_4 + c_5 + 4 c_6 S: | 3 c_2 = c_4 Sb: | 2 c_2 = 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 = 32/3 c_2 = 1 c_3 = 4/3 c_4 = 3 c_5 = 32/3 c_6 = 2 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 32 c_2 = 3 c_3 = 4 c_4 = 9 c_5 = 32 c_6 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 32 HNO_3 + 3 Sb_2S_3 ⟶ 4 H_2O + 9 H_2SO_4 + 32 NO + 6 HSbO4
Balance the chemical equation algebraically: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Sb_2S_3 ⟶ c_3 H_2O + c_4 H_2SO_4 + c_5 NO + c_6 HSbO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, S and Sb: H: | c_1 = 2 c_3 + 2 c_4 + c_6 N: | c_1 = c_5 O: | 3 c_1 = c_3 + 4 c_4 + c_5 + 4 c_6 S: | 3 c_2 = c_4 Sb: | 2 c_2 = 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 = 32/3 c_2 = 1 c_3 = 4/3 c_4 = 3 c_5 = 32/3 c_6 = 2 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 32 c_2 = 3 c_3 = 4 c_4 = 9 c_5 = 32 c_6 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 32 HNO_3 + 3 Sb_2S_3 ⟶ 4 H_2O + 9 H_2SO_4 + 32 NO + 6 HSbO4

Structures

+ ⟶ + + + HSbO4
+ ⟶ + + + HSbO4

Names

nitric acid + antimony(III) sulfide ⟶ water + sulfuric acid + nitric oxide + HSbO4
nitric acid + antimony(III) sulfide ⟶ water + sulfuric acid + nitric oxide + HSbO4

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO4 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: 32 HNO_3 + 3 Sb_2S_3 ⟶ 4 H_2O + 9 H_2SO_4 + 32 NO + 6 HSbO4 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 | 32 | -32 Sb_2S_3 | 3 | -3 H_2O | 4 | 4 H_2SO_4 | 9 | 9 NO | 32 | 32 HSbO4 | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 32 | -32 | ([HNO3])^(-32) Sb_2S_3 | 3 | -3 | ([Sb2S3])^(-3) H_2O | 4 | 4 | ([H2O])^4 H_2SO_4 | 9 | 9 | ([H2SO4])^9 NO | 32 | 32 | ([NO])^32 HSbO4 | 6 | 6 | ([HSbO4])^6 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])^(-32) ([Sb2S3])^(-3) ([H2O])^4 ([H2SO4])^9 ([NO])^32 ([HSbO4])^6 = (([H2O])^4 ([H2SO4])^9 ([NO])^32 ([HSbO4])^6)/(([HNO3])^32 ([Sb2S3])^3)
Construct the equilibrium constant, K, expression for: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO4 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: 32 HNO_3 + 3 Sb_2S_3 ⟶ 4 H_2O + 9 H_2SO_4 + 32 NO + 6 HSbO4 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 | 32 | -32 Sb_2S_3 | 3 | -3 H_2O | 4 | 4 H_2SO_4 | 9 | 9 NO | 32 | 32 HSbO4 | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 32 | -32 | ([HNO3])^(-32) Sb_2S_3 | 3 | -3 | ([Sb2S3])^(-3) H_2O | 4 | 4 | ([H2O])^4 H_2SO_4 | 9 | 9 | ([H2SO4])^9 NO | 32 | 32 | ([NO])^32 HSbO4 | 6 | 6 | ([HSbO4])^6 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])^(-32) ([Sb2S3])^(-3) ([H2O])^4 ([H2SO4])^9 ([NO])^32 ([HSbO4])^6 = (([H2O])^4 ([H2SO4])^9 ([NO])^32 ([HSbO4])^6)/(([HNO3])^32 ([Sb2S3])^3)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO4 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: 32 HNO_3 + 3 Sb_2S_3 ⟶ 4 H_2O + 9 H_2SO_4 + 32 NO + 6 HSbO4 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 | 32 | -32 Sb_2S_3 | 3 | -3 H_2O | 4 | 4 H_2SO_4 | 9 | 9 NO | 32 | 32 HSbO4 | 6 | 6 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 | 32 | -32 | -1/32 (Δ[HNO3])/(Δt) Sb_2S_3 | 3 | -3 | -1/3 (Δ[Sb2S3])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) H_2SO_4 | 9 | 9 | 1/9 (Δ[H2SO4])/(Δt) NO | 32 | 32 | 1/32 (Δ[NO])/(Δt) HSbO4 | 6 | 6 | 1/6 (Δ[HSbO4])/(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 = -1/32 (Δ[HNO3])/(Δt) = -1/3 (Δ[Sb2S3])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/9 (Δ[H2SO4])/(Δt) = 1/32 (Δ[NO])/(Δt) = 1/6 (Δ[HSbO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO4 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: 32 HNO_3 + 3 Sb_2S_3 ⟶ 4 H_2O + 9 H_2SO_4 + 32 NO + 6 HSbO4 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 | 32 | -32 Sb_2S_3 | 3 | -3 H_2O | 4 | 4 H_2SO_4 | 9 | 9 NO | 32 | 32 HSbO4 | 6 | 6 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 | 32 | -32 | -1/32 (Δ[HNO3])/(Δt) Sb_2S_3 | 3 | -3 | -1/3 (Δ[Sb2S3])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) H_2SO_4 | 9 | 9 | 1/9 (Δ[H2SO4])/(Δt) NO | 32 | 32 | 1/32 (Δ[NO])/(Δt) HSbO4 | 6 | 6 | 1/6 (Δ[HSbO4])/(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 = -1/32 (Δ[HNO3])/(Δt) = -1/3 (Δ[Sb2S3])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/9 (Δ[H2SO4])/(Δt) = 1/32 (Δ[NO])/(Δt) = 1/6 (Δ[HSbO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | antimony(III) sulfide | water | sulfuric acid | nitric oxide | HSbO4 formula | HNO_3 | Sb_2S_3 | H_2O | H_2SO_4 | NO | HSbO4 Hill formula | HNO_3 | S_3Sb_2 | H_2O | H_2O_4S | NO | HO4Sb name | nitric acid | antimony(III) sulfide | water | sulfuric acid | nitric oxide | IUPAC name | nitric acid | thioxo-(thioxostibanylthio)stibane | water | sulfuric acid | nitric oxide |
| nitric acid | antimony(III) sulfide | water | sulfuric acid | nitric oxide | HSbO4 formula | HNO_3 | Sb_2S_3 | H_2O | H_2SO_4 | NO | HSbO4 Hill formula | HNO_3 | S_3Sb_2 | H_2O | H_2O_4S | NO | HO4Sb name | nitric acid | antimony(III) sulfide | water | sulfuric acid | nitric oxide | IUPAC name | nitric acid | thioxo-(thioxostibanylthio)stibane | water | sulfuric acid | nitric oxide |

Random Queries

HCl + KNO3 = HNO3 + KCl
freezing point of sodium hydroxide vs lithium hydroxide
MDL number of 1,3-propylene glycol
CAS number of heavy molecules
chemical types of propionitrile-d 5 vs 1-chloro-2-propanol
specific gravity of hafnium phosphide vs titanium(II) bromide
trans-10,11-dibromodibenzosuberone
1,1,1,3,3,3-hexafluoro-2-propan(ol-d)
telluronium cation
alternate names of cobalt aluminum oxide