Following reaction has a reaction yield of #75%#. In order to obtain #"25 g"# of carbon dioxide, what is the necessary amount of propane in moles? #"C"_3"H"_8 + "O"_2 -> "CO"_2 + "H"_2"O"# (unbalanced)

1 Answer
Dec 24, 2017

Answer:

#"0.25 moles C"_3"H"_8#

Explanation:

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

#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((l))#

Now, the balanced chemical equation tells you that in order for the reaction to produce #3# moles of carbon dioxide, it must consume #1# mole of propane.

This is the theoretical yield of the reaction, i.e. what you would get at #100%# yield.

In your case, the reaction is known to have a percent yield equal to #75%#. This essentially means that for every #100# moles of carbon dioxide that the reaction could theoretically produce, you only get #75# moles.

Use the molar mass of carbon dioxide to convert the mass of the sample to moles

#25 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.568 moles CO"_2#

Now, at #100%# yield, the reaction must consume

#0.568 color(red)(cancel(color(black)("moles CO"_2))) * ("1 mole C"_3"H"_8)/(3color(red)(cancel(color(black)("moles CO"_2)))) = "0.189 moles C"_3"H"_8#

in order to produce #0.568# moles of carbon dioxide. However, you know that the reaction has a percent yield equal to #75%#, which means that the reaction produced only #75%# of the number of moles of carbon dioxide that it would have produced at theoretical yield.

This means that in order to get #0.568# moles of carbon dioxide, the reaction must consume enough moles of propane to theoretically produce

#0.568 color(red)(cancel(color(black)("moles CO"_2))) * overbrace("100 moles CO"_2/(75 color(red)(cancel(color(black)("moles CO"_2)))))^(color(blue)("= 75% yield")) = "0.757 moles CO"_2#

You can thus say that the number of moles of propane needed is equal to

#0.757 color(red)(cancel(color(black)("moles CO"_2))) * ("1 mole C"_3"H"_8)/(3color(red)(cancel(color(black)("moles CO"_2)))) = color(darkgreen)(ul(color(black)("0.25 moles C"_3"H"_8)))#

The answer is rounded to two sig figs, the number of sig figs you have for the mass of carbon dioxide.

So, you can say that, for this reaction, you have

#"0.189 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 100% yield")aaa))(->) " " "0.568 moles CO"_2#

#"0.25 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 100% yield")aaa))(->) " " "0.757 moles CO"_2#

#"0.25 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 75% yield")aaa))(->) " " "0.568 moles CO"_2#