Question #60166

The first thing to do here is figure out the volume of oxygen gas present in the cylinder by using the percent concentration by volume of air.

Since air is said to be #21%# by volume oxygen gas, #"O"_2#, you can say that you get #"21 mL"# of oxygen gas for every #"100 mL"# of air. The cylinder will thus contain

#506.2 color(red)(cancel(color(black)("mL air"))) * "21 mL O"_2/(100color(red)(cancel(color(black)("mL air")))) = "106.3 mL O"_2#

Now, take a look at the balanced chemical equation that describes the combustion of octane, #"C"_8"H"_18#

#color(red)(2)"C"_ 8"H"_ (18(l)) + color(blue)(25)"O"_ (2(g)) -> 16"CO"_ (2(g)) + 18"H"_ 2"O"_((l))#

Notice that the reaction consumes #color(blue)(25)# moles of oxygen gas for every #color(red)(2)# moles of octane. Your job now will be to determine how many moles of oxygen gas you have in the cylinder.

To do that, use the ideal gas law equation, which looks like this

#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

Before plugging in your values, make sure that your units for pressure, temperature, and volume match those used in the expression of the universal gas constant.

Use the following conversion factors

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm " = " 760 mmHg")color(white)(a/a)|))) "" # #" "color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^3"mL")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)|)))#

Rearrange the ideal gas law equation to solve for #n#, the number of moles of oxygen gas present in the cylinder

#PV = nRT implies n =(PV)/(RT)#

Plug in your values to find

#n = (752.7/760 color(red)(cancel(color(black)("atm"))) * 106.3 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (55.2 + 273.15)color(red)(cancel(color(black)("K"))))#

#n = "0.0039054 moles O"_2#

In order for all the moles of oxygen gas to react, you need to have

#0.0039054 color(red)(cancel(color(black)("moles O"_2))) * (color(red)(2)color(white)(a)"moles C"_8"H"_18)/(color(blue)(25)color(red)(cancel(color(black)("moles O"_2)))) = "0.00031243 moles C"_8"H"_18#

Use octane's molar mass to determine how many grams would be equivalent to this many moles of octane

#0.00031243 color(red)(cancel(color(black)("moles C"_8"H"_18))) * "114.2285 g"/(1color(red)(cancel(color(black)("mole C"_8"H"_18)))) = color(green)(|bar(ul(color(white)(a/a)"0.0357 g"color(white)(a/a)|)))#

The answer is rounded to three sig figs.