# Question 4c71e

Mar 14, 2016

#### Answer:

$\text{7.618 g}$

#### Explanation:

Potassium chlorate, ${\text{KClO}}_{3}$, will decompose upon heating to form potassium chloride, $\text{KCl}$, and oxygen gas, ${\text{O}}_{2}$, according to the following balanced chemical equation

$\textcolor{b l u e}{2} {\text{KClO"_text(3(s]) stackrel(color(red)(Delta)color(white)(aa))(->) 2"KCl"_text((s]) + color(purple)(3)"O}}_{\textrm{2 \left(g\right]}}$ $\uparrow$

The reaction will produce $\textcolor{p u r p \le}{3}$ moles of oxygen gas for every $\textcolor{b l u e}{2}$ moles of potassium chlorate that decompose.

This will be the theoretical yield of the reaction, i.e. what you get when the reaction has a 100% yield.

So, if absolutely every mole of potassium chlorate that decomposes upon heating ends up producing $\frac{\textcolor{p u r p \le}{3}}{\textcolor{b l u e}{2}}$ moles of oxygen gas, then the reaction is said to have a 100% yield $\to$ this will be your theoretical yield.

Use potassium chlorate's molar mass to determine how many moles you have in that $\text{19.45-g}$ sample

19.45color(red)(cancel(color(black)("g"))) * "1 mole KClO"_3/(122.55color(red)(cancel(color(black)("g")))) = "0.15871 moles KClO"_3

This means that theoretically, the reaction should produce

0.15871color(red)(cancel(color(black)("moles KClO"_3))) * (color(purple)(3)color(white)(a)"moles O"_2)/(color(blue)(2)color(red)(cancel(color(black)("moles KClO"_3)))) = "0.23807 moles O"_2#

To determine how many grams of oxygen would contain this many moles, use oxygen gas' molar mass

$0.23807 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)(|bar(ul(color(white)(a/a)"7.618 g O}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to four sig figs, the number of sig figs you have for the mass of potassium chlorate.

If your reaction ends up producing less oxygen than predicted, you can calculate its percent yield by using

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% yield" = "what you actually get"/"what you should theoretically get} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, for example, if you end up with $\text{5.555 g}$ of oxygen gas, you can say that the reaction had a percent yield of

$\text{% yield" = (5.555 color(red)(cancel(color(black)("g"))))/(7.618color(red)(cancel(color(black)("g")))) xx 100 = "72.92%}$