What is the limiting reagent when #12.5*lb# pentane are reacted with a #12.5*lb# mass of dioxygen gas?

Question

Olivia Anton · Accepted Answer

We need (i) a stoichiometric equation:

#C_5H_12 + 8O_2 rarr 5CO_2(g) + 6H_2O# And (ii) we need equivalent quantities of dioxygen and pentane. We know that #"1 lb"# #-=# #454*g#: #"Moles of pentane"=(12.5*lbxx454*g*lb^-1)/(72.15*g*mol^-1)=78.7*mol# #"Moles of dioxygen"=(12.5*lbxx454*g*lb^-1)/(32.0*g*mol^-1)=177.3*mol# Clearly, there is INSUFFICIENT dioxygen gas for complete combustion. And thus #O_2# is the limiting reagent. Complete combustion requires #8xx78.7*mol# dioxygen gas. What is this as a mass? The question also suggested mixing pentane and oxygen, which is not something I would be comfortable doing and would make me flee quickly if I witnessed someone doing it.

Jeremiah Adams · Answer

The limiting reactant is oxygen.

Here's an additional method for determining the limiting reactant. We are aware that the masses, moles, and molar masses of the constituent compounds must be included in a balanced equation. 1. Compile all the data in one location, placing the molar masses above the formulas and the rest below them. #M_r:color(white)(mmmmm) 72.15color(white)(mm)32.00# #color(white)(mmmmmmm)"C"_5"H"_12 +color(white)(ll) "8O"_2 → "5CO"_2 + "6H"_2"O"# #"Amt/lb-mol:"color(white)(ll)0.1733color(white)(ml)0.3906# #"Divide by:"color(white)(mmml)1color(white)(mmmm)8# #"Moles rxn:"color(white)(mll)0.1733color(white)(ml)"0.048 83"# Note: Since the numbers will remain in the same ratio, we do not have to use gram-moles; instead, we can use pound-moles. #"Moles of C"_5"H"_12 = 12.5 color(red)(cancel(color(black)("lb C"_5"H"_12))) × ("1 lb-mol C"_5"H"_12)/(72.15 color(red)(cancel(color(black)("lb C"_5"H"_12)))) = "0.1733 lb-mol C"_5"H"_12# #"Moles of O"_2 = 12.5 color(red)(cancel(color(black)("lb O"_2))) × ("1 lb-mol O"_2)/(32.00 color(red)(cancel(color(black)("lb O"_2)))) = "0.3906 lb-mol O"_2# 2. Determine which reactant is limiting. Finding the "moles of reaction" that each will yield will make it simple to determine which reactant is the limiting one. In the balanced equation, you divide the moles of each reactant by the corresponding coefficient. In the table above, I completed that for you. #"O"_2# is the limiting reactant because it gives the fewest moles of reaction.

Jose Ayala · Answer

undefined To determine the limiting reagent, we need to compare the moles of each reactant. First, we convert the masses of pentane and dioxygen gas to moles using their respective molar masses. The molar mass of pentane (C5H12) is approximately 72.15 g/mol, and the molar mass of dioxygen gas (O2) is approximately 32 g/mol.

12.5 lb pentane * (453.592 g / 1 lb) / (72.15 g/mol) = 78.18 mol pentane

12.5 lb dioxygen gas * (453.592 g / 1 lb) / (32 g/mol) = 177.34 mol dioxygen gas

Since there are fewer moles of pentane compared to dioxygen gas, pentane is the limiting reagent.