Final answer:
The calculation shows that 54.0 kg of Al2O3 will be produced when 28.6 kg of Al reacts with an excess of Fe2O3 in a thermite reaction. This answer is not among the provided options, indicating a potential error in the question or choices.
Step-by-step explanation:
The question asks how many kilograms of aluminum oxide (Al2O3) will be produced if 28.6 kg of aluminum (Al) reacts with an excess of iron(III) oxide (Fe2O3) in a thermite reaction. This is a stoichiometry problem involving molar mass and mole-to-mole conversion. To solve this, we need to follow these steps
Step 1: Moles of Al = 28600 g / 26.98 g/mol = 1060.04 mol Al
Step 2: The molar ratio of Al to Al2O3 is 2:1, so for every 2 moles of Al, 1 mole of Al2O3 is produced.
Step 3: Moles of Al2O3 = 1060.04 mol Al / 2 = 530.02 mol Al2O3
Step 4: Mass of Al2O3 = 530.02 mol * 101.96 g/mol = 54023.5 g
Step 5: Mass of Al2O3 in kg = 54023.5 g / 1000 = 54.0 kg
The correct answer is not provided in the options a, b, c, and d. There might be a mistake in the question or answer choices. Based on calculations, 54.0 kg of Al2O3 will be produced.