Aluminum oxide can be reduced to aluminum metal using carbon, the other reaction product being carbon monoxide. Determine the enthalpy change when 12.5 g of aluminum is produced by this method. [$\Delta$H^$\circ$_f(carbon monoxide) = -110.5 kJ/mol; $\Delta$H^$\circ$_f(aluminum oxide) = -1669.8 kJ/mol]

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The enthalpy change for the reaction can be calculated using the following equation:$$\Delta H = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)}$$where $\Delta H_f$ is the enthalpy of formation of the substance.For the reaction:$$2Al_2O_3 + 3C \rightarrow 4Al + 3CO$$The enthalpy change is:$$\Delta H = [4\Delta H_f(Al) + 3\Delta H_f(CO)] - [2\Delta H_f(Al_2O_3) + 3\Delta H_f(C)]$$Substituting the given values:$$\Delta H = [4(0) + 3(-110.5)] - [2(-1669.8) + 3(0)]$$$$\Delta H = -331.5 + 3339.6$$$$\Delta H = 3008.1 kJ$$To determine the enthalpy change when 12.5 g of aluminum is produced, we need to convert the mass of aluminum to moles and then multiply by the enthalpy change per mole:$$12.5 g Al \times \frac{1 mol Al}{26.98 g Al} \times 3008.1 kJ/mol = 361 kJ$$Therefore, the enthalpy change when 12.5 g of aluminum is produced is 361 kJ.

Aluminum Oxide Can Be Reduced to Aluminum Metal Using Carbon Δ\DeltaΔ

Aluminum Oxide Can Be Reduced to Aluminum Metal Using Carbon $\Delta$