A common laboratory preparation for small for quantities of #O_2# is to decompose #KClO_3# by heating: #2KClO_3 (s) -> 2KCl (s) + 3O_2 (g)#. If the decomposition of 2.00 g of #KCIO_3# gives 0.720 g of #O_2#, what is the percent yield for the reaction?

1 Answer
Sep 1, 2016

Answer:

#91.9%#

Explanation:

Start by examining the balanced chemical equation that describes this decomposition reaction

#color(darkgreen)(2)"KClO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"KCl"_ ((s)) + color(blue)(3)"O"_ (2(g))#

Notice that it takes #color(darkgreen)(2)# moles of potassium chlorate to produce #color(blue)(3)# moles of oxygen gas. This represents the reaction's theoretical yield, i.e. the amount of oxygen gas that you can expect to see for a reaction that has a #100%# yield.

The reaction provides you with grams of potassium chlorate, so use the molar masses of potassium chlorate and of oxygen gas to convert the mole ratio to a gram ratio.

#(color(darkgreen)(2)color(red)(cancel(color(black)("moles KClO"_3))) * "122.55 g"/(1color(red)(cancel(color(black)("mole KClO"_3)))))/(color(blue)(3)color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2))))) = "2.553 g KClO"_3/"1 g O"_2#

This means that for every #"2.553 g"# of potassium chlorate that undergo combustion, the reaction can theoretically produce #"1 g"# of oxygen gas.

Now, the problem tells you that you decompose #"2.00 g"# of potassium chlorate. The reaction should theoretically produce

#2.00 color(red)(cancel(color(black)("g KClO"_3))) * "1 g O"_2/(2.553color(red)(cancel(color(black)("g KClO"_3)))) = "0.7834 g O"_2#

This represents the reaction's theoretical yield. You know that the reaction actually produced #"0.720 g"# of oxygen gas, which means that its percent yield was

#(0.720 color(red)(cancel(color(black)("g O"_2))))/(0.7834color(red)(cancel(color(black)("g O"_2))))xx 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(91.9%)color(white)(a/a)|)))#

The answer is rounded to three sig figs.