Fe(NO3)3 + H2CO3 = HNO3 + Fe2(CO3)3

Fe(NO_3)_3 ferric nitrate + H_2CO_3 carbonic acid ⟶ HNO_3 nitric acid + Fe2(CO3)3

Balance the chemical equation algebraically: Fe(NO_3)_3 + H_2CO_3 ⟶ HNO_3 + Fe2(CO3)3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe(NO_3)_3 + c_2 H_2CO_3 ⟶ c_3 HNO_3 + c_4 Fe2(CO3)3 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, N, O, C and H: Fe: | c_1 = 2 c_4 N: | 3 c_1 = c_3 O: | 9 c_1 + 3 c_2 = 3 c_3 + 9 c_4 C: | c_2 = 3 c_4 H: | 2 c_2 = c_3 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 6 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Fe(NO_3)_3 + 3 H_2CO_3 ⟶ 6 HNO_3 + Fe2(CO3)3

+ ⟶ + Fe2(CO3)3

ferric nitrate + carbonic acid ⟶ nitric acid + Fe2(CO3)3

K_c = ([HNO3]^6 [Fe2(CO3)3])/([Fe(NO3)3]^2 [H2CO3]^3)

rate = -1/2 (Δ[Fe(NO3)3])/(Δt) = -1/3 (Δ[H2CO3])/(Δt) = 1/6 (Δ[HNO3])/(Δt) = (Δ[Fe2(CO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| ferric nitrate | carbonic acid | nitric acid | Fe2(CO3)3 formula | Fe(NO_3)_3 | H_2CO_3 | HNO_3 | Fe2(CO3)3 Hill formula | FeN_3O_9 | CH_2O_3 | HNO_3 | C3Fe2O9 name | ferric nitrate | carbonic acid | nitric acid | IUPAC name | iron(+3) cation trinitrate | carbonic acid | nitric acid |

| ferric nitrate | carbonic acid | nitric acid | Fe2(CO3)3 molar mass | 241.86 g/mol | 62.024 g/mol | 63.012 g/mol | 291.71 g/mol phase | solid (at STP) | | liquid (at STP) | melting point | 35 °C | | -41.6 °C | boiling point | | | 83 °C | density | 1.7 g/cm^3 | | 1.5129 g/cm^3 | solubility in water | very soluble | | miscible | dynamic viscosity | | | 7.6×10^-4 Pa s (at 25 °C) |