Fe2O3 = O2 + Fe

iron(III) oxide ⟶ oxygen + iron

Balance the chemical equation algebraically: ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 ⟶ c_2 + c_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe and O: Fe: | 2 c_1 = c_3 O: | 3 c_1 = 2 c_2 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3/2 c_3 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 ⟶ 3 + 4

⟶ +

iron(III) oxide ⟶ oxygen + iron

| iron(III) oxide | oxygen | iron molecular enthalpy | -826 kJ/mol | 0 kJ/mol | 0 kJ/mol total enthalpy | -1652 kJ/mol | 0 kJ/mol | 0 kJ/mol | H_initial = -1652 kJ/mol | H_final = 0 kJ/mol | ΔH_rxn^0 | 0 kJ/mol - -1652 kJ/mol = 1652 kJ/mol (endothermic) | |

| iron(III) oxide | oxygen | iron molecular entropy | 90 J/(mol K) | 205 J/(mol K) | 27 J/(mol K) total entropy | 180 J/(mol K) | 615 J/(mol K) | 108 J/(mol K) | S_initial = 180 J/(mol K) | S_final = 723 J/(mol K) | ΔS_rxn^0 | 723 J/(mol K) - 180 J/(mol K) = 543 J/(mol K) (endoentropic) | |

| iron(III) oxide | oxygen | iron Hill formula | Fe_2O_3 | O_2 | Fe name | iron(III) oxide | oxygen | iron IUPAC name | | molecular oxygen | iron

| iron(III) oxide | oxygen | iron molar mass | 159.69 g/mol | 31.998 g/mol | 55.845 g/mol phase | solid (at STP) | gas (at STP) | solid (at STP) melting point | 1565 °C | -218 °C | 1535 °C boiling point | | -183 °C | 2750 °C density | 5.26 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 7.874 g/cm^3 solubility in water | insoluble | | insoluble surface tension | | 0.01347 N/m | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | odor | odorless | odorless |