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

Apr 9, 2017

We need (i) a stoichiometric equation:

${C}_{5} {H}_{12} + 8 {O}_{2} \rightarrow 5 C {O}_{2} \left(g\right) + 6 {H}_{2} O$

#### Explanation:

And (ii) we need equivalent quantities of dioxygen and pentane. We know that $\text{1 lb}$ $\equiv$ $454 \cdot g$:

$\text{Moles of pentane} = \frac{12.5 \cdot l b \times 454 \cdot g \cdot l {b}^{-} 1}{72.15 \cdot g \cdot m o {l}^{-} 1} = 78.7 \cdot m o l$

$\text{Moles of dioxygen} = \frac{12.5 \cdot l b \times 454 \cdot g \cdot l {b}^{-} 1}{32.0 \cdot g \cdot m o {l}^{-} 1} = 177.3 \cdot m o l$

Clearly, there is INSUFFICIENT dioxygen gas for complete combustion. And thus ${O}_{2}$ is the limiting reagent. Complete combustion requires $8 \times 78.7 \cdot m o l$ dioxygen gas. What is this as a mass?

By the way the question proposed that oxygen and pentane were mixed together.........this is not something that I would be happy doing, and I would run a long way away if I saw someone doing this....

Apr 9, 2017

The limiting reactant is oxygen.

#### Explanation:

Here's another way to identify the limiting reactant.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and everything else below the formulas.

${M}_{r} : \textcolor{w h i t e}{m m m m m} 72.15 \textcolor{w h i t e}{m m} 32.00$
$\textcolor{w h i t e}{m m m m m m m} \text{C"_5"H"_12 +color(white)(ll) "8O"_2 → "5CO"_2 + "6H"_2"O}$
$\text{Amt/lb-mol:} \textcolor{w h i t e}{l l} 0.1733 \textcolor{w h i t e}{m l} 0.3906$
$\text{Divide by:} \textcolor{w h i t e}{m m m l} 1 \textcolor{w h i t e}{m m m m} 8$
$\text{Moles rxn:"color(white)(mll)0.1733color(white)(ml)"0.048 83}$

Note: We do not have to stick with gram-moles. We can use pound-moles because the numbers will still be in the same ratio.

${\text{Moles of C"_5"H"_12 = 12.5 color(red)(cancel(color(black)("lb C"_5"H"_12))) × ("1 lb-mol C"_5"H"_12)/(72.15 color(red)(cancel(color(black)("lb C"_5"H"_12)))) = "0.1733 lb-mol C"_5"H}}_{12}$

${\text{Moles of O"_2 = 12.5 color(red)(cancel(color(black)("lb O"_2))) × ("1 lb-mol O"_2)/(32.00 color(red)(cancel(color(black)("lb O"_2)))) = "0.3906 lb-mol O}}_{2}$

2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

${\text{O}}_{2}$ is the limiting reactant because it gives the fewest moles of reaction.