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
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
This is the theoretical yield of the reaction, i.e. what you would get at
In your case, the reaction is known to have a percent yield equal to
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
#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
This means that in order to get
#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#