Answer:
458.8 grams of Al2O3 will be produced. The limiting reactant is Al
Explanation:
Step 1: Data given
Mass of Al = 243 grams
Atomic mass Al = 26.98 g/mol
STP = 1 atm and 273 K
Step 2: The balanced equation
4 Al + 3 O2 → 2 Al2O3
Step 3: Calculate moles Al
Moles Al = mass Al / molar mass Al
Moles Al = 243 grams / 26.98 g/mol
Moles Al = 9.00 moles
Step 4: Calculate moles O2
22.4 L = 1 mol
211 L = 9.42 moles
Step 4: Calculate moles the limiting reatant
For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3
Al is the limiting reactant. It will completely be consumed ( 9.00moles). O2 is in excess. There will react 3/4 * 9.00 = 6.75 moles
There will remain 9.42 - 6.75 = 2.67 moles
Step 5: Calculate mass Al2O3
For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3
Mass Al2O3 = moles Al2O3 * molar mass Al2O3
Mass Al2O3 = 4.50 moles * 101.96 g/mol
Mass Al2O3 = 458.8 grams
458.8 grams of Al2O3 will be produced. The limiting reactant is Al
