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
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
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
Now, the problem tells you that you decompose
#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 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.