Combustion Reaction of Acetylene: Finding the Limiting Reactant

Which reactant is the limiting reactant in the combustion reaction for acetylene?

Consider the combustion reaction for acetylene:
2C2H2(l) + 5O2(g) -> 4CO2(g) + 2H2O(g)
If the acetylene tank contains 37.0 mol of C2H2 and the oxygen tank contains 81.0 mol of O2, what is the limiting reactant for this reaction?

Answer: Oxygen

The combustion reaction for acetylene can be represented as:
2C2H2(l) + 5O2(g) -> 4CO2(g) + 2H2O(g)

From the balanced chemical equation, we can see that 2 moles of acetylene react with 5 moles of oxygen. Therefore, 37 moles of acetylene will react with (5/2) x 37 = 92.5 moles of oxygen. However, the available amount of oxygen is only 81.0 moles.

Looking at the reaction the other way, 5 moles of oxygen react with 2 moles of acetylene. So, 81 moles of oxygen will react with (2/5) x 81 = 32.4 moles of acetylene. This means that oxygen is the limiting reactant in this reaction.

The limiting reactant is the reactant that limits the formation of the product. In this case, oxygen is the limiting reactant, while acetylene is the excess reactant as there are 4.6 moles of acetylene left unreacted.