Question

A sample of Al2O3 decomposed to pure aluminum and oxygen. After decomposition, 29.0 moles of Al were recovered. How many moles of Al2O3 were decomposed?

Answer 1

To produce 29.0 moles of Al, 14.5 moles of Al2O3 must be decomposed according to the stoichiometry of the reaction which is based on the molar ratio of 2 moles Al to 1 mole Al2O3.

Step-by-step explanation:

To determine how many moles of Al2O3 were decomposed to produce 29.0 moles of Al, we must first look at the stoichiometry of the decomposition reaction:

Al2O3(s) → 2 Al(s) + 3/2 O2(g)

From the balanced equation, we see that 1 mole of Al2O3 yields 2 moles of Al. Therefore, to produce 29.0 moles of Al, we would need half as many moles of Al2O3, which is 14.5 moles of Al2O3. This calculation is based on the molar ratio between Al and Al2O3 from the balanced equation: 2 moles Al: 1 mole Al2O3.

Answer 2

The decomposition of Al2O3 to form pure aluminum (Al) along with the release of oxygen (O2) gas can be represented by the following chemical reaction:

2Al2O3 → 4Al + 3O2

Based on the reaction stoichiometry:

2 moles of aluminum oxide decomposes to form 4 moles of pure aluminum

Therefore, if 29.0 moles of aluminum is recovered, then the moles of aluminum oxide required will be:

= 29.0 moles Al * 2 moles Al2O3/4 moles Al

= 14.5 moles of Al2O3