How many grams of Al2O3 are produced when 211 liters of O2 reacts with 243 grams of Al at STP?

4 Al + 3 O2 → 2 Al2O3 What is the limiting reactant?

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