Al + CuO = Cu + Al2O3

aluminum + cupric oxide ⟶ copper + aluminum oxide

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 Al, Cu and O: Al: | c_1 = 2 c_4 Cu: | c_2 = c_3 O: | c_2 = 3 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 + 3 ⟶ 3 +

+ ⟶ +

aluminum + cupric oxide ⟶ copper + aluminum oxide

| aluminum | cupric oxide | copper | aluminum oxide molecular enthalpy | 0 kJ/mol | -157.3 kJ/mol | 0 kJ/mol | -1676 kJ/mol total enthalpy | 0 kJ/mol | -471.9 kJ/mol | 0 kJ/mol | -1676 kJ/mol | H_initial = -471.9 kJ/mol | | H_final = -1676 kJ/mol | ΔH_rxn^0 | -1676 kJ/mol - -471.9 kJ/mol = -1204 kJ/mol (exothermic) | | |

| aluminum | cupric oxide | copper | aluminum oxide molecular entropy | 28.3 J/(mol K) | 43 J/(mol K) | 33 J/(mol K) | 51 J/(mol K) total entropy | 56.6 J/(mol K) | 129 J/(mol K) | 99 J/(mol K) | 51 J/(mol K) | S_initial = 185.6 J/(mol K) | | S_final = 150 J/(mol K) | ΔS_rxn^0 | 150 J/(mol K) - 185.6 J/(mol K) = -35.6 J/(mol K) (exoentropic) | | |

| aluminum | cupric oxide | copper | aluminum oxide Hill formula | Al | CuO | Cu | Al_2O_3 name | aluminum | cupric oxide | copper | aluminum oxide IUPAC name | aluminum | | copper | dialuminum;oxygen(2-)