Question

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)

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Answer #1

#95%#

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 #5# moles of oxygen gas for every #1# mole of propane that takes part in the reaction.

This means that, in order for the reaction to consume #7# moles of propane, it must also 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 #7# moles of oxygen gas, you can say that oxygen gas will be the limiting reagent, i.e. it will be completely consumed before all the moles of propane will get the chance to react.

So the reaction will consume #7# moles of oxygen gas and

#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 #100%# yield

#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 #4.1# moles of carbon dioxide, so you can say that the percent yield of the reaction was

#"% 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.

answered by: Stefan V.
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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?
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