If you start with 13.4g of c3h8 and 4.35g of o2 determine the limiting reagent

Balance chemical equation for this reaction is,

                           C₃H₈  +  10 O₂    →     3 CO₂   +   4H₂O

According to this equation 44 g of C₃H₈ reacts with 320 g of O₂ to completely consume.

Now let's calculate amount of O₂ required to completely consume 13.4 g of C₃H₈,

44 g C₂H₈ was completely consumed by reacting  =  320 g of O₂
13.4 g of C₂H₈ will require                                       =  X g of O₂

Solving for X,
                      X  =  (13.4 g × 320 g) ÷ 44 g

                      X  =  97.45 g of O₂ is required to consume 13.4 g of C₂H₈
Now,
Let's calculate amount of C₂H₈ required to consume 4.35 g of O₂,

320 g of O₂ was consumed by reacting  =  44 g of C₂H₈
4.35 g of O₂ will require                          =  X g of C₂H₈

Solving for X,
                     X  =  (4.35 g ×44 g) ÷ 320

                     X  =  0.598 g of C₂H₈ is required to consume 4.35 g of O₂

Result:
           As the given amount of C₂H₈ requires 97.45 g of O₂, but we are provided only with 4.35 g, it means the oxygen will consume first.

           So, Oxygen is the limiting reagent.

True, the standard unit for volume is liters which can be represented by either a lower case "l" or a capital "l"