If you start with 7.07.0 moles of propane and 7.07.0 moles of oxygen gas what is the percent yield if 4.04.0 moles of carbon dioxide are produced?
Consider the reaction below. If you start with 7.0 moles of C3H8 (propane) and 7.0 moles of O2, _ is the percent yield if 4.0 moles of carbon dioxide is produced.
C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)
Consider the reaction below. If you start with 7.0 moles of C3H8 (propane) and 7.0 moles of O2, _ is the percent yield if 4.0 moles of carbon dioxide is produced.
C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)
1 Answer
Explanation:
The first thing that you need to do here is to figure out the theoretical yield of the reaction.
"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((g))C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
You know that the reaction consumes
This means that, in order for the reaction to consume
7.0 color(red)(cancel(color(black)("moles C"_3"H"_8))) * "5 moles O"_2/(1color(red)(cancel(color(black)("moles C"_3"H"_8)))) = "35 moles O"_2
Since you have only
So the reaction will consume
7.0 color(red)(cancel(color(black)("moles O"_2))) * ("1 mole C"_3"H"_8)/(5color(red)(cancel(color(black)("moles O"_2)))) = "1.4 moles C"_3"H"_8
In this case, you can say that the reaction can theoretically produce, i.e. what you would get at
7 color(red)(cancel(color(black)("moles O"_2))) * "3 moles CO"_2/(5color(red)(cancel(color(black)("moles O"_2)))) = "4.2 moles CO"_2
However, you know that the actual yield of the reaction is
"% yield" = (4.0 color(red)(cancel(color(black)("moles CO"_2))))/(4.2color(red)(cancel(color(black)("moles CO"_2)))) * 100% = color(darkgreen)(ul(color(black)(95%)))
The answer is rounded to two sig figs.