Hello, Mohamed,

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

The balanced equation tells us we need 5 moles O_{2} for every 1 mole of C_{3}H_{8}, a molar ratio of 5 to 1. If we have 5 moles of propane, we would need 25 moles of O_{2}. We only have 10, so oxygen is the limiting reagent.

Let's assume all 10 moles of O_{2} are consumed and determine how much propane will remain. With a ratio of 5/1, 10 moles of O_{2} will consume 2 moles of C_{3}H_{8}. That leaves 3 moles propane unreacted. 3 moles times propane's molar mass of 44 g/mole, means 132.1 grams of propane remains, if grams is tthe desired unit.

Bob