# 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?

Sep 1, 2016

91.9%

#### Explanation:

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

$\textcolor{\mathrm{da} r k g r e e n}{2} {\text{KClO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"KCl"_ ((s)) + color(blue)(3)"O}}_{2 \left(g\right)}$

Notice that it takes $\textcolor{\mathrm{da} r k g r e e n}{2}$ moles of potassium chlorate to produce $\textcolor{b l u e}{3}$ 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 $\text{2.553 g}$ of potassium chlorate that undergo combustion, the reaction can theoretically produce $\text{1 g}$ of oxygen gas.

Now, the problem tells you that you decompose $\text{2.00 g}$ 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 $\text{0.720 g}$ 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.