Final answer:
The mass of aluminum oxide (Al2O3) produced from burning 108 g of aluminum in oxygen is 203.92 g, determined by stoichiometric calculations using the molar masses of aluminum and aluminum oxide.
Step-by-step explanation:
The question asks what mass of aluminum oxide is produced when 108 g of aluminum is burned in oxygen, following the reaction equation: 4Al + 3O2 → 2Al2O3. To find the mass of aluminum oxide produced, we must first calculate the number of moles of aluminum by using its molar mass, which is 26.98 g/mol. This calculation is done as follows:
Number of moles of Al = mass of Al (g) / molar mass of Al (g/mol)
Number of moles of Al = 108 g / 26.98 g/mol ≈ 4 moles of Al
According to the stoichiometry of the reaction, 4 moles of aluminum (Al) produce 2 moles of aluminum oxide (Al2O3). Since we have 4 moles of Al available, all will react to produce exactly 2 moles of Al2O3. The molar mass of Al2O3 is 101.96 g/mol, so we can now find the mass of aluminum oxide produced by multiplying the number of moles of Al2O3 by its molar mass:
Mass of Al2O3 = number of moles of Al2O3 × molar mass of Al2O3
Mass of Al2O3 = 2 moles × 101.96 g/mol = 203.92 g
Therefore, the mass of aluminum oxide produced when 108 g of aluminum is burned in oxygen is 203.92 g.