# What volume of O_2, measured 82.1°C and 713 mm Hg, will be produced by the decomposition of 5.71 g of KClO_3?

Nov 15, 2016

The reaction produces ${\text{2.18 L of O}}_{2}$.

#### Explanation:

Step 1. Write the balanced chemical equation.

The balanced equation is

${\text{2KClO"_3 → "2KCl" + "3O}}_{2}$

Step 2. The Procedure

The problem is to convert grams of ${\text{KClO}}_{3}$ to moles of ${\text{O}}_{2}$ and volume of ${\text{O}}_{2}$.

The procedure is:

(a) Use the molar mass to convert the mass of ${\text{KClO}}_{3}$ to moles of ${\text{KClO}}_{3}$.

(b) Use the molar ratio (from the balanced equation) to convert moles of ${\text{KClO}}_{3}$ to moles of ${\text{O}}_{2}$.

(e) Use the Ideal Gas Law to convert moles of ${\text{O}}_{2}$ to volume of ${\text{O}}_{2}$.

In equation form,

${\text{grams of KClO"_3 stackrelcolor(blue)("molar mass"color(white)(ml)) (→) "moles of KClO"_3 stackrelcolor(blue)("molar ratio"color(white)(ml))→ "moles of O"_2 stackrelcolor(blue)("Ideal Gas Law"color(white)(ml))(→) "volume of O}}_{2}$

The Calculations

(a) Moles of ${\text{KClO}}_{3}$

5.71 color(red)(cancel(color(black)("g KClO"_3))) × ("1 mol KClO"_3)/(122.55 color(red)(cancel(color(black)("g KClO"_3)))) = "0.046 59 mol KClO"_3

(b) Moles of ${\text{O}}_{2}$

${\text{0.046 59"color(red)(cancel(color(black)("mol KClO"_3))) × ("3 mol O"_2)/(2 color(red)(cancel(color(black)("mol KClO"_3)))) = "0.069 89 mol O}}_{2}$

(c) Volume of ${\text{O}}_{2}$

The Ideal Gas Law is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this to give

$V = \frac{n R T}{P}$

$n = {\text{0.069 89 mol O}}_{2}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(82.1 + 273.15) K" = "355.25 K}$
P = 711 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.9355 atm"

V = ("0.069 89" color(red)(cancel(color(black)("mol"))) × "0.082 06 L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 355.25color(red)(cancel(color(black)( "K"))))/(0.9355 color(red)(cancel(color(black)("atm")))) = "2.18 L"

The volume of ${\text{O}}_{2}$ produced is $\text{2.18 L}$.