Answer:
1.96 × 10^23 aluminum atoms
Step-by-step explanation:
To determine the number of aluminum atoms that can be produced by the decomposition of 33.3 g of Al2O3, we need to use the balanced chemical equation for the decomposition reaction of Al2O3:
2 Al2O3 -> 4 Al + 3 O2
From this equation, we can see that for every 2 moles of Al2O3, 4 moles of aluminum are produced. We can use this stoichiometric relationship to convert the mass of Al2O3 to moles of aluminum.
First, we need to calculate the molar mass of Al2O3:
2 Al + 3 O -> Al2O3
2(26.98 g/mol) + 3(16.00 g/mol) = 101.96 g/mol
So, 1 mole of Al2O3 has a mass of 101.96 g.
Now, we can use the mass of Al2O3 given in the problem to calculate the number of moles of aluminum that can be produced:
moles of Al2O3 = 33.3 g / 101.96 g/mol = 0.326 mol
moles of Al = 2/2 * moles of Al2O3 = 0.326 mol
Finally, we can convert the number of moles of aluminum to the number of aluminum atoms:
number of Al atoms = moles of Al * Avogadro's number
number of Al atoms = 0.326 mol * 6.022 × 10^23 atoms/mol = 1.96 × 10^23 atoms
Therefore, approximately 1.96 × 10^23 aluminum atoms can be produced by the decomposition of 33.3 g of Al2O3.