# Question 186eb

Mar 26, 2016

${\text{13.1 g O}}_{2}$

#### Explanation:

Start by writing the balanced chemical equation that describes the decomposition of potassium chlorate, ${\text{KClO}}_{3}$, to form potassium chloride, $\text{KCl}$, and oxygen gas, ${\text{O}}_{2}$

$2 {\text{KClO"_text(3(s]) -> color(red)(2)"KCl"_text((s]) + color(blue)(3)"O}}_{\textrm{2 \left(g\right]}} \uparrow$

Notice that for every two moles of potassium chlorate that undergo decomposition, you get $\textcolor{red}{2}$ moles of potassium chloride and $\textcolor{b l u e}{3}$ moles of oxygen gas.

You can convert this $\textcolor{red}{2} : \textcolor{b l u e}{3}$ mole ratio that exists between the two products of the reaction to a gram ratio by using the molar masses of the two compounds.

${\text{For KCl": " " M_M = "74.55 g mol}}^{- 1}$

${\text{For O"_2: " " M_M = "32.0 g mol}}^{- 1}$

This tells you that every mole of potassium chloride has a mass of $\text{74.55 g}$, which implies that $\textcolor{red}{2}$ moles will have a mass of

2 color(red)(cancel(color(black)("moles KCl"))) * "74.55 g"/(1color(red)(cancel(color(black)("mole KCl")))) = "149.1 g"

For oxygen gas, every mole has a mass of $\text{32.0 g}$, which means that $\textcolor{b l u e}{3}$ moles will have a mass of

3 color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = "96.0 g"

The $\textcolor{red}{2} : \textcolor{b l u e}{3}$ mole ratio will be equivalent to a $149.1 : 96.0$ gram ratio, i.e. you get $\text{96.0 g}$ of oxygen gas for every $\text{149.1 g}$ of potassium chloride produced by the reaction.

This means that $\text{20.3 g}$ of potassium chloride would correspond to

20.3color(red)(cancel(color(black)("g KCl"))) * "96.0 g O"_2/(149.1color(red)(cancel(color(black)("g KCl")))) = "13.07 g O"_2#

Rounded to three sig figs, the number of sig figs you have for the mass of potassium chloride, the answer will be

${m}_{{O}_{2}} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{13.1 g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$