Question #5d6cd

1 Answer
Mar 17, 2016


Here's what I got.


The theoretical yield of a chemical reaction is the amount of product that you should expect to get if the reaction had a #100%# yield.

In other words, the theoretical yield will tell you how much product will be formed if the reaction ends up forming products according to the stoichiometric coefficients that exist between the reactants and the products in the balanced chemical equation.

In your case, you have

#"Xe"_text((g]) + color(red)(2)"F"_text(2(g]) -> "XeF"_text(4(s])#

Notice that the reaction consumes #color(red)(2)# moles of fluorine gas for every #1# mole of xenon that takes part in the reaction.

Likewise, the reaction produces #1# mole of xenon tetrafluoride for every #1# mole of xenon that gets consumed.

Now, in order to determine the theoretical yield of the reaction, you need to figure out how many moles of each reactant you have present in the reaction vessel.

To do that, use the molar masses of xenon and fluorine gas to go from grams to moles

#130. color(red)(cancel(color(black)("g"))) * "1 mole Xe"/(131.293color(red)(cancel(color(black)("g")))) = "0.99015 moles Xe"#

#100. color(red)(cancel(color(black)("g"))) * "1 mole F"_2/(37.997color(red)(cancel(color(black)("g")))) = "2.6318 moles F"_2#

So, do you have enough fluorine gas to ensure that all the moles of xenon react?

Use the #1:color(red)(2)# mole ratio that exists between the two reactants to figure that out.

#0.99015 color(red)(cancel(color(black)("moles Xe"))) * (color(red)(2)color(white)(a)"moles F"_2)/(1color(red)(cancel(color(black)("mole Xe")))) = "1.9803 moles F"_2#

In order for all the moles of xenon to react, you need to provide #1.9803# moles of fluorine gas. Since you have more than you need, you can say that xenon will act as a limiting reagent, i.e. it will determine how much fluorine gas reacts.

In other words, fluorine gas will be in excess here.

Now, if #0.99015# moles of xenon take part in the reaction, you can theoretically get

#0.99015 color(red)(cancel(color(black)("moles Xe"))) * "1 mole XeF"_4/(1color(red)(cancel(color(black)("mole Xe")))) = "0.99015 moles XeF"_4#

Use the molar mass of xenon tetrafluoride to determine how many grams would contain this many moles

#0.99015color(red)(cancel(color(black)("moles XeF"_4))) * "207.287 g"/(1color(red)(cancel(color(black)("mole XeF"_4)))) = color(green)(|bar(ul(color(white)(a/a)"205 g"color(white)(a/a)|)))#

This will be your reaction's theoretical yield.

Notice that the reaction ended up producing #"145 g"# of xenon tetrafluoride. This represents the reaction's actual yield, i.e. what you actually get by doing the reaction in the lab.

Since percent yield is defined as

#color(blue)(|bar(ul(color(white)(a/a)"% yield" = "what you actually get"/"what you should theoretically get" xx 100 color(white)(a/a)|)))#

you can say that this reaction has a percent yield of

#"% yield" = (145 color(red)(cancel(color(black)("g"))))/(205color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"70.7%"color(white)(a/a)|)))#

Both answers are rounded to three sig figs, the number of sig figs you have for the masses of the two reactants.