# Question #35151

Mar 5, 2015

Your reaction will produce $\text{22.5 g}$ of carbon dioxide.

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

The most important tool in your arsenal when dealing with stoichiometry problems is the mole ratio. The mole ratio tells you in what exact proportions the reactans will mix and exactly how much products their reaction will produce.

In this case, the mole ratio between oxygen and carbon dioxide is $\text{5:3}$, i.e. you need 5 moles of oxygen to produce 3 moles of carbon dioxide.

Simply put, the number of carbon dioxide moles produced will always be exactly 3/5 times smaller than the number of oxygen moles that react.

An important aspect is that you have excess propane, which means that oxygen will be the limiting reagent $\to$ all the moles of oxygen will react.

Determine how many moles of oxygen you have by using its molar mass

$\text{27.3 g" * "1 mole"/"32.0 g" = "0.853 moles oxygen}$

Now use the aforementioned mole ratio to determine how many moles of $C {O}_{2}$ will be produced

${\text{0.853 moles oxygen" * "3 moles carbon dioxide"/"5 moles oxygen" = "0.512 moles CO}}_{2}$

Finally, use carbon dioxide's molar mass to determine the mass produced

$\text{0.512 moles" * "44.0 g"/"1 mole" = "22.5 g carbon dioxide}$