The major source of aluminum in the world this bauxite (mostly aluminum oxide). It’s thermal decomposition can be represented by: Al2 O3 (s) —> 2 Al (s) + 3/2 O2 (g)

ΔH rxn = 1676

If aluminum is produced this way, how many grams of aluminum can conform when 1.000×10^3 kJ of heat is transferred?

The correct answer is 32.2 grams.

Explanation:

Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,  

ΔHrxn = 1676/2 = 838 kJ/mol

Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,  

(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum

The grams of aluminum produced can be obtained by using the formula,  

mass = moles * molecular mass

= 1.19 * 26.98

= 32.2 grams.