# In a closed chamber, 58 grams of propane gas (C3H8) and 200 grams of oxygen are sparked to initiate the comubustion what is the limited reagent in this reaction?

Mar 9, 2018

We write the stoichiometric equation....

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right) + \Delta$

#### Explanation:

And then we interrogate the molar quantities....

$\text{Moles of propane} \equiv \frac{58.0 \cdot g}{44.10 \cdot g \cdot m o {l}^{-} 1} = 1.31 \cdot m o l .$

$\text{Moles of dioxygen} \equiv \frac{200.0 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} = 6.25 \cdot m o l .$

But CLEARLY, we require $6.55 \cdot m o l$ of dioxygen for stoichiometric equivalence... And thus dioxygen is the LIMITING REAGENT for the reaction as written....

In practice, the hydrocarbon could combust incompletely, to give say $C \left(s\right)$, i.e. soot....or carbon monoxide...$C O \left(g\right)$...and we could represent this incomplete combustion by the reaction...

${C}_{3} {H}_{8} \left(g\right) + \frac{7}{2} {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + C O \left(g\right) + C \left(s\right) + 4 {H}_{2} O \left(l\right)$

How do we know how much carbon and carbon monoxide we get? How else but by analysis of the of the product mix...? And clearly the terms of the question do not provide these data.

Mar 9, 2018

Oxygen is the limited reagent

#### Explanation:

The reaction equation is:

${C}_{3} {H}_{8} + 5 {O}_{2} = 3 C {O}_{2} + 4 {H}_{2} O + H e a t$

Molar mass of Propane is 44.09 gms/mol
Molar mass of Oxygen is 31.99 gms/mol
58gms of propane is 1.314 moles , i.e ( $\frac{58}{44.09}$)

From above equation you need 15 moles of oxygen to complete burm 1 mole of Propane.

So, 1.314 mole of Propane requires 6.58 moles of oxygenwhich would weigh 210 gms.

But the chamber is closed and you have only 200gms of oxygen, which means all the Propane did not burn . You needed more Oxygen.

Oxygen is the limited reagent