Final answer:
To find the mass of oxygen required to react with 5.3960 grams of aluminium, one must first calculate the moles of aluminium, then use stoichiometry to find the corresponding moles of oxygen, and finally convert those moles into grams, resulting in 4.800 grams of oxygen needed.
Step-by-step explanation:
To calculate the mass of oxygen required to react completely with 5.3960 grams of aluminium, we must use the concept of stoichiometry. The reaction between aluminium (Al) and oxygen (O2) to form aluminium oxide (Al2O3) can be represented by the balanced chemical equation:
4 Al + 3 O2 → 2 Al2O3
This equation indicates that 4 moles of Al react with 3 moles of O2 to produce 2 moles of Al2O3. To find how many moles of Al there are in 5.3960 grams, we divide the mass by the molar mass of aluminium (approximately 26.98 g/mol):
Moles of Al = 5.3960 g Al / 26.98 g/mol Al = 0.200 mol Al
Next, we use the mole ratio from the balanced chemical equation to find the moles of O2 required:
Moles of O2 required = (0.200 mol Al) × (3 mol O2 / 4 mol Al) = 0.150 mol O2
Finally, to find the mass of O2, we multiply the moles by the molar mass of O2 (approximately 32.00 g/mol):
Mass of O2 required = 0.150 mol O2 × 32.00 g/mol O2 = 4.800 g O2
Therefore, 4.800 grams of oxygen are required to react completely with 5.3960 grams of aluminium.