# 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?

##### 1 Answer

#### Answer:

#### Explanation:

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

#"C"_ 2"H"_ (6(g)) + 7/2"O"_ (2(g)) -> color(blue)(2)"CO"_ (2(g)) + 3"H"_ 2"O"_ ((l))#

Notice that every **mole** of ethane that undergoes combustion produces **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

#13.0^@"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**

#color(blue)(ul(color(black)(PV = nRT)))#

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 theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof the gas

Rearrange to solve for

#PV = nRT implies V = (nRT)/P#

and plug in your values to find

#V = (29.26 color(red)(cancel(color(black)("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"))))#

#V = color(darkgreen)(ul(color(black)("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