Question #06eb8

COMPLETE LAB SHEET

https://docs.google.com/viewer?a=v&pid=sites&srcid=Y2ZzZDE2Lm9yZ3xtcy1iaWxsc3xneDoxNDg5OGMwZjNkYWY3M2Zj

This answer is based on the solution provided here:

http://socratic.org/questions/2-what-volume-of-oxygen-must-react-with-a-full-tank-of-gas-for-the-following-at-

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 #color(red)(2):16# mole ratio that exists between octane and carbon dioxide to determine how many 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

#P# - the pressure of the gas
#V# - the volume it occupies
#n# - the number of moles of gas
#R# - the universal gas constant, usually given as #0.0821("atm" * "L")/("mol" * "K")#
#T# - the 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 #V#, the volume of the gas

#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!