If you start with #7.0# moles of propane and #7.0# moles of oxygen gas what is the percent yield if #4.0# 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))#
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.