How many mL of O_2 gas at 25°C and 755 mm Hg pressure can be produced from the thermal decomposition of 0.300 grams of KClO_3?

Jun 22, 2016

The reaction will produce 82.9 mL of ${\text{O}}_{2}$.

Explanation:

Step 1. Write the balanced chemical equation.

The balanced equation is

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

Step 2. Strategy

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

We can use the flow chart below to help us. The process is:

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

0.300 color(red)(cancel(color(black)("g KClO"_3))) × ("1 mol KClO"_3)/( 122.55 color(red)(cancel(color(black)("g KClO"_3)))) = "0.002 448 mol KClO"_3

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

0.002 448color(red)(cancel(color(black)("mol KClO"_3))) × ("3 mol O"_2)/(2 color(red)(cancel(color(black)("mol KClO"_3)))) = "0.003 672 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

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

$n = \text{0.003 672 mol}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(25 + 273.15) K" = "298.15 K}$

P = 755 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.9934 atm"

V = ("0.003 672" color(red)(cancel(color(black)("mol"))) × "0.082 06"color(white)(l) "L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 273.15color(red)(cancel(color(black)( "K"))))/(0.9934 color(red)(cancel(color(black)("atm")))) = "0.0829 L" = " 82.9 mL"

The volume of ${\text{O}}_{2}$ produced is $\text{82.9 mL}$.