What is the limiting reactant when 3 grams of propane is mixed with 10 grams of Oxygen?

C3H8 + O2 —> H2O + CO2

The answer to your question is Propane

Explanation:

Data

mass of Propane = 3 g

mass of Oxygen = 10 g

limiting reactant = ?

Balanced chemical reaction

               C₃H₈  +  O₂  ⇒   H₂O  +  CO₂

Process

1.- Calculate the molar mass of the reactants

C₃H₈ = (12 x 3) + (8 x 1) = 36 + 8 = 44 g

O₂ = 16 x 2 = 32 g

2.- Calculate the theoretical yield  C₃H₈ / O₂ = 44/32 = 1.375

3.- Calculate the experimental yield C₃H₈/O₂ = 3/10 = 0.3

4.- Conclusion

The limiting reactant is Propane because the experimental yield was lower than the theoretical yield.

1. they vary with a higher number because boiling is way hotter than melting