Answer to Question #235200 in General Chemistry for NHLAKANIPHO

Question

If 255.0g of Octane and 1510.0g of Oxygen are present, determine which reagent is limiting

Answer 1

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O.

Molar mass of Octane =114.23

255/114.23 = 2.23moles

Molar mass of Oxygen = 15.999

1510/15.999 = 94.381 moles

2 mol of octane is allowed to react with 25 mol of oxygen. Oxygen is the limiting reactant.