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.

The correct answer is "atomic weight".