HNO3 + Fe2O3 = H2O + Fe(NO3)3

nitric acid + iron(III) oxide ⟶ water + ferric nitrate

Balance the chemical equation algebraically: + ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Fe: H: | c_1 = 2 c_3 N: | c_1 = 3 c_4 O: | 3 c_1 + 3 c_2 = c_3 + 9 c_4 Fe: | 2 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 = 6 c_2 = 1 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 + ⟶ 3 + 2

+ ⟶ +

nitric acid + iron(III) oxide ⟶ water + ferric nitrate

K_c = ([H2O]^3 [Fe(NO3)3]^2)/([HNO3]^6 [Fe2O3])

rate = -1/6 (Δ[HNO3])/(Δt) = -(Δ[Fe2O3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/2 (Δ[Fe(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | iron(III) oxide | water | ferric nitrate Hill formula | HNO_3 | Fe_2O_3 | H_2O | FeN_3O_9 name | nitric acid | iron(III) oxide | water | ferric nitrate IUPAC name | nitric acid | | water | iron(+3) cation trinitrate

| nitric acid | iron(III) oxide | water | ferric nitrate molar mass | 63.012 g/mol | 159.69 g/mol | 18.015 g/mol | 241.86 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) melting point | -41.6 °C | 1565 °C | 0 °C | 35 °C boiling point | 83 °C | | 99.9839 °C | density | 1.5129 g/cm^3 | 5.26 g/cm^3 | 1 g/cm^3 | 1.7 g/cm^3 solubility in water | miscible | insoluble | | very 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) | odor | | odorless | odorless |