Final answer:
To find the amount of aluminum that remained unreacted, convert the given mass of aluminum to moles, subtract the moles of aluminum sulfate formed from the moles of aluminum reacted, and convert the remaining moles of aluminum back to grams. The correct answer is 0 grams.
Step-by-step explanation:
To determine the amount of aluminum that remained unreacted, we need to find the amount of aluminum that reacted with the sulfuric acid. From the balanced chemical equation, we know that 2 moles of aluminum react with 3 moles of sulfuric acid to produce 3 moles of hydrogen gas and 1 mole of aluminum sulfate.
First, we need to convert the given mass of aluminum (80.0 grams) to moles. The molar mass of aluminum is 26.98 g/mol, so:
Moles of aluminum = Mass of aluminum / Molar mass of aluminum
Moles of aluminum = 80.0 g / 26.98 g/mol = 2.964 mol
From the balanced chemical equation, we can see that the moles of aluminum sulfate formed will be the same as the moles of aluminum reacted. Therefore, the amount of aluminum remaining unreacted is:
Aluminum remaining unreacted = Moles of aluminum - Moles of aluminum sulfate formed = 2.964 mol - 2.964 mol = 0 mol
Therefore, the correct answer is a) 0 grams.