# Question #06eb8

##### 1 Answer

Here's what I got.

#### Explanation:

**COMPLETE LAB SHEET**

This answer is based on the solution provided here:

So, you know that the balanced chemical equation for this combustion reaction looks like this

#color(red)(2)"C"_8"H"_text(18(l]) + color(blue)(25)"O"_text(2(g]) -> 16"CO"_text(2(g]) + 18"H"_2"O"_text((l])#

You also know that the reaction used up

#18.5color(red)(cancel(color(black)("gal"))) * (3.79color(blue)(cancel(color(black)("L"))))/(1color(red)(cancel(color(black)("gal")))) * (1000color(green)(cancel(color(black)("mL"))))/(1color(blue)(cancel(color(black)("L")))) * "0.703 g"/(1color(green)(cancel(color(black)("mL")))) = 4.93 * 10^4"g"#

of octane. This many grams of octane will contain

#4.93 * 10^4color(red)(cancel(color(black)("g"))) * "1 mole octane"/(114.23color(red)(cancel(color(black)("g")))) = "431.6 moles octane"#

Now, your job at this point is to use the *moles* of carbon dioxide will be produced by the reaction.

#431.6 color(red)(cancel(color(black)("moles octane"))) * "16 moles CO"_2/(color(red)(2)color(red)(cancel(color(black)("moles octane")))) = "3452.8 moles CO"_2#

The problem provides you with the pressure and temperature inside the combustion chamber, which means that you can use the **ideal gas law** equation to find the volume occupied by the carbon dioxide

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "# , where

*universal gas constant*, usually given as

**absolute temperature** of the gas

Convert the temperature from *degrees Fahrenheit* to *degrees Celsius*, then from *degrees Celsius* to *Kelvin* by using the conversion factors

#color(purple)(|bar(ul(color(white)(a/a)color(black)(t[""^@"C"] = (t[""^@"F"] - 32) xx 5/9)color(white)(a/a)|))) " "# and#" "color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#

You should get

#t[""^@"C"] = (1940^@"F" - 32) * 5/9 = 1060^@"C"#

and

#T["K"] = 1060^@"C" + 273.15 = "1333.15 K"#

Rearrange the ideal gas law equation to solve for

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

Plug in your values to get

#V = (3452.8color(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")))) * 1333.15color(red)(cancel(color(black)("K"))))/(11.56color(red)(cancel(color(black)("atm"))))#

#V = color(green)(|bar(ul(color(white)(a/a)"32,700 L"color(white)(a/a)|))) -># rounded to three sig figs

Make sure to double-check my calculations!