# Question 94034

Feb 12, 2017

The mass of potassium chlorate = 34.8 g

#### Explanation:

${\text{2KClO"_3 rarr "2KCl" + "3O}}_{2} \textcolor{w h i t e}{m m m m m m m m m m m m} \left(1\right)$

$\text{Volume O"_2 = "10.0 L}$
$T = \text{(150 + 273.15) K" = "423.15 K}$
P = "150 kPa" = "1.48 atm"color(white)(m)("1 kPa = 0.00986 atm")

According to the Ideal Gas equation,

$P V = n R T$
$R = {\text{0.08206 L"."atm.K"^(-1)."mol}}^{- 1}$
$n = {\text{no. of moles of O}}_{2}$

$P V = n R T$

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

n = ("1.48 atm" xx "10.0 L")/("0.08206 L·atm"·"K"^(-1)·"mol"^(-1) xx 423.15 K) = "0.4262 mol"

$\text{O"_2 = "0.4262 mol} \textcolor{w h i t e}{m m m m m m m m m m m m m m m m} \left(2\right)$

According to the balanced equation

3 mol of ${\text{O}}_{2}$ is produced from the decomposition of 2 mol of ${\text{KClO}}_{3}$

0.4262 mol of ${\text{O}}_{2}$ will be produced from

$\frac{2 \times 0.4262}{3} \textcolor{w h i t e}{l} {\text{mol KClO"_3 = "0.2841 mol of KClO}}_{3}$

Therefore $\text{KClO"_3 = "0.2841 mol}$

$\text{molar mass of KClO"_3 = "122.55 g/mol}$

"mass of KClO"_3 = "122.55 g/mol" xx "0.2841 mol" = bb("34.82 g")

Feb 12, 2017

The reaction requires 34.8 g of ${\text{KClO}}_{3}$.

#### Explanation:

There are four steps involved in this problem:

1. Write the balanced equation for the reaction.
2. Use the Ideal Gas Law to calculate the moles of ${\text{O}}_{2}$.
3. Use the molar ratio of ${\text{KClO"_3:"O}}_{2}$ from the balanced equation to calculate the moles of ${\text{KClO}}_{3}$.
4. Use the molar mass of ${\text{KClO}}_{3}$ to calculate the mass of ${\text{KClO}}_{3}$.

Let's get started.

Step 1. Write the balanced chemical equation.

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

Step 2. Calculate the moles of ${\text{O}}_{2}$.

The Ideal Gas Law is

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

$P = \text{150 kPa}$
$V = \text{10.0 L}$
$R = \text{8.314 kPa·L·K"^"-1""mol"^"-1}$
$T = \text{150 °C" = "423.15 K}$

We can rearrange the Ideal Gas Law to get:

n = (PV)/(RT) = (150color(red)(cancel(color(black)("kPa"))) × 10.0 color(red)(cancel(color(black)("L"))))/("8.314"color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 423.15 color(red)(cancel(color(black)("K")))) = "0.4623 mol"#

3. Calculate the moles of ${\text{KClO}}_{3}$.

${\text{Moles of KClO"_3 = "0.4264" color(red)(cancel(color(black)("mol O"_2))) × "2 mol KClO"_3/(3 color(red)(cancel(color(black)("mol O"_2)))) = "0.2842 mol KClO}}_{3}$

4. Calculate the mass of ${\text{KClO}}_{3}$

${\text{Mass of KClO"_3 = 0.2842 color(red)(cancel(color(black)("mol KClO"_3))) × "122.55 g KClO"_3/(1 color(red)(cancel(color(black)("mol KClO"_3)))) ="34.8 g KClO}}_{3}$