(Stoichiometry)

Question

1 C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g) - Identify the Limiting Reactant in the following problem and identify how much of the excess reactant remains after the reaction has reached completion. If you try to react 5 moles of Propane (C₃H₈) with 10 moles of O₂, which reactant will you run out of first? How much of the excess reactant will you have leftover?

Answer 1

Hello, Mohamed,

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

The balanced equation tells us we need 5 moles O₂ for every 1 mole of C₃H₈, a molar ratio of 5 to 1. If we have 5 moles of propane, we would need 25 moles of O₂. We only have 10, so oxygen is the limiting reagent.

Let's assume all 10 moles of O₂ are consumed and determine how much propane will remain. With a ratio of 5/1, 10 moles of O₂ will consume 2 moles of C₃H₈. 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

Answer 2

From the balanced equation, we see that we need 5 moles of O₂ for each 1 mole of propane. So, if you have 5 moles of propane, you will need 25 moles of O₂. This can be calculated as follows:

5 moles propane x 5 moles O₂ / 1 mol propane = 25 moles O₂.

Since you have only 10 moles of O₂, the O₂ will run out first.

How much excess (propane) is left? Again, this can be calculated as follows:

10 moles O₂ x 1 mol propane / 5 moles O₂ = 2 mols propane used up or reacted

Since you started with 5 moles of propane, the amount left will be 5 - 2 = 3 mols propane left over.

Hope this makes sense.