# Question ef2ef

Aug 17, 2015

This reaction will produce 6272 L of carbon dioxide under those conditions of pressure and temperature.

#### Explanation:

I assume that the problem wants you to treat oxygen as being in excess, which is another way of saying that you are to assume that all the moles of propanol will take part in the reaction.

The balanced chemical equation for this combustion reaction is

$\textcolor{red}{2} {C}_{3} {H}_{7} O {H}_{\left(l\right)} + 9 {O}_{2 \left(g\right)} \to \textcolor{b l u e}{6} C {O}_{2 \left(g\right)} + 8 {H}_{2} {O}_{\left(l\right)}$

Notice that you have a $\textcolor{red}{1} : \textcolor{b l u e}{3}$ ($2 : 6$) mole ratio between propanol and carbon dioxide. This means that, regardless of how many moles of propanol react, the reaction will always produce three times as many moles of carbon dioxide.

In other words, if you know how many moles of propanol reacted, you know how many moles of carbon dioxide were produced. To determine how many moles of propanol reacted, use the compound's molar mass

6011.0color(red)(cancel(color(black)("g"))) * "1 mole propanol"/(60.1color(red)(cancel(color(black)("g")))) = "100.02 moles propanol"

The reaction produced

100.02color(red)(cancel(color(black)("moles propanol"))) * (color(blue)(3)" moles CO"""_2)/(color(red)(1)color(red)(cancel(color(black)("mole propanol")))) = "300.06 moles CO"""_2#

To figure out what volume of carbon dioxide would contain this many moles at ${20}^{\circ} \text{C}$ and $\text{1.150 atm}$, use the ideal gas law equation

$P V = n R T \implies V = \frac{n R T}{P}$

${V}_{C {O}_{2}} = \left(300.06 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mole"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 20)color(red)(cancel(color(black)("K"))))/(1.150color(red)(cancel(color(black)("atm}}}}\right)$
${V}_{C {O}_{2}} = \text{6272.1 L}$
${V}_{C {O}_{2}} = \textcolor{g r e e n}{\text{6272 L}}$