# Suppose 0.440 kg of ethane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0 °C. What is the volume of carbon dioxide gas that is produced?

Jun 10, 2017

$\text{690 L}$

#### Explanation:

Start by writing a balanced chemical equation that describes this combustion reaction

${\text{C"_ 2"H"_ (6(g)) + 7/2"O"_ (2(g)) -> color(blue)(2)"CO"_ (2(g)) + 3"H"_ 2"O}}_{\left(l\right)}$

Notice that every $1$ mole of ethane that undergoes combustion produces $\textcolor{b l u e}{2}$ moles of carbon dioxide, so start by calculating the number of moles of ethane present in your sample.

To do that, use the molar mass of ethane

0.440 color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * ("1 mole C"_2"H"_6)/(30.07color(red)(cancel(color(black)("g")))) = "14.63 moles C"_2"H"_6

This means that the reaction will produce

14.63 color(red)(cancel(color(black)("moles C"_2"H"_6))) * (color(blue)(2)color(white)(.)"moles CO"_2)/(1color(red)(cancel(color(black)("mole C"_2"H"_6)))) = "29.26 moles CO"_2

Now, you know that the carbon dioxide is collected at a pressure of $\text{1 atm}$ and a temperature of

${13.0}^{\circ} \text{C" = 13.0^@"C" + 273.15 = "286.15 K}$

At this point, you can calculate the volume of carbon dioxide produced by the reaction by using the ideal gas law equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Rearrange to solve for $V$

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

and plug in your values to find

$V = \left(29.26 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0/0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 286.15color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm}}}}\right)$

$V = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{690 L}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the pressure of the gas, which implies that the answer should be $\text{700 L}$.