HNO3 + FeS = H2O + SO2 + NO + Fe(NO3)3

HNO_3 nitric acid + FeS ferrous sulfide ⟶ H_2O water + SO_2 sulfur dioxide + NO nitric oxide + Fe(NO_3)_3 ferric nitrate

Balance the chemical equation algebraically: HNO_3 + FeS ⟶ H_2O + SO_2 + NO + Fe(NO_3)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 FeS ⟶ c_3 H_2O + c_4 SO_2 + c_5 NO + c_6 Fe(NO_3)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Fe and S: H: | c_1 = 2 c_3 N: | c_1 = c_5 + 3 c_6 O: | 3 c_1 = c_3 + 2 c_4 + c_5 + 9 c_6 Fe: | c_2 = c_6 S: | c_2 = c_4 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 = 16/3 c_2 = 1 c_3 = 8/3 c_4 = 1 c_5 = 7/3 c_6 = 1 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 16 c_2 = 3 c_3 = 8 c_4 = 3 c_5 = 7 c_6 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 16 HNO_3 + 3 FeS ⟶ 8 H_2O + 3 SO_2 + 7 NO + 3 Fe(NO_3)_3

+ ⟶ + + +

nitric acid + ferrous sulfide ⟶ water + sulfur dioxide + nitric oxide + ferric nitrate

K_c = ([H2O]^8 [SO2]^3 [NO]^7 [Fe(NO3)3]^3)/([HNO3]^16 [FeS]^3)

rate = -1/16 (Δ[HNO3])/(Δt) = -1/3 (Δ[FeS])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/3 (Δ[SO2])/(Δt) = 1/7 (Δ[NO])/(Δt) = 1/3 (Δ[Fe(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | ferrous sulfide | water | sulfur dioxide | nitric oxide | ferric nitrate formula | HNO_3 | FeS | H_2O | SO_2 | NO | Fe(NO_3)_3 Hill formula | HNO_3 | FeS | H_2O | O_2S | NO | FeN_3O_9 name | nitric acid | ferrous sulfide | water | sulfur dioxide | nitric oxide | ferric nitrate IUPAC name | nitric acid | | water | sulfur dioxide | nitric oxide | iron(+3) cation trinitrate