Answer:
The minimum mass of octane that could be left over is 43.0 grams
Explanation:
Step 1: Data given
Mass of octane = 73.0 grams
Mass of oxygen = 105.0 grams
Molar mass octane = 114.23 g/mol
Molar mass oxygen = 32.0 g/mol
Step 2: The balanced equation
2C8H18 + 25O2 → 16CO2 + 18H2O
Step 3: Calculate the number of moles
Moles = mass / molar mass
Moles octane = 73.0 grams / 114.23 g/mol
Moles octane = 0.639 moles
Moles O2 = 105.0 grams / 32.0 g/mol
Moles O2 = 3.28 moles
Step 4: Calculate the limiting reactant
For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O
O2 is the limiting reactant. It will completely be consumed. (3.28 moles). There will react 3.28 / 12.5 = 0.2624 moles. There will remain 0.639 - 0.2624 = 0.3766 moles octane
Step 5: Calculate mass octane remaining
Mass octane = moles * molar mass
Mass octane = 0.3766 moles * 114.23 g/mol
Mass octane = 43.0 grams
The minimum mass of octane that could be left over is 43.0 grams
