Hi, Joselyne,

The "secret" to solving this problem is recognizing that the ratios provided in the equation are CONSTANT (meaning that for this particular reaction, those ratios will always be as shown here):

C_{3}H_{8}(g) + 5O_{2}(g) → 3CO_{2}(g) + 4H_{2}O(g)

So, the ratio of propane to oxygen is (and will be!) 1:5. Therefore, if you are performing a reaction with 3 moles of propane and 10 moles of oxygen, you can solve for the limiting reagent as follows:

With 3 moles of propane, the overall molar quantities used/produced would be:

**3**×1 C_{3}H_{8}(g) + **3**×5 O_{2}(g) → **3**×3 CO_{2}(g) + **3**×4 H_{2}O(g) ,

or **3** C_{3}H_{8}(g) + **15** O_{2}(g) → **9 **CO_{2}(g) + **12** H_{2}O(g).

As you can see, to fully consume **3** moles of propane would require 15 moles of oxygen to satisfy the combustion equation ... but you don't *have *15 moles of O_{2}(g), you only have 10! Therefore, **oxygen is the limiting reagent in this reaction**.

If you want to check your results, we can solve the same equation starting with the 10 moles of oxygen. In this case:

2×1 C_{3}H_{8}(g) + 2×5 O_{2}(g) → 2×3 CO_{2}(g) + 2×4 H_{2}O(g)

**2 ** C_{3}H_{8}(g) + **10** O_{2}(g) → 6 CO_{2}(g) + 8 H_{2}O(g)

Here, you see that 10 moles of oxygen would consume 2 moles of propane ... and there would still be one mole of propane remaining (in excess). **The limiting reagent** - the one to be exhausted __first __- **would be oxygen.**

Thanks,

Chris

Joselyne L.

Thank you so much for your help!04/27/20