Question #94034

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Question #94034

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Answer 1 · 2017-02-12T03:58:15

The mass of potassium chlorate = 34.8 g

Explanation:

${2KClO}_{3} \to 2KCl + {3O}_{2} m m m m m m m m m m m m (1)$ $"2KClO"_3 rarr "2KCl" + "3O"_2 color(white)(mmmmmmmmmmmm)(1)$

${Volume O}_{2} = 10.0 L$ $"Volume O"_2 = "10.0 L"$
$T = (150 + 273.15) K = 423.15 K$ $T = "(150 + 273.15) K" = "423.15 K"$
$P = 150 kPa = 1.48 atm m (1 kPa = 0.00986 atm)$ $P = "150 kPa" = "1.48 atm"color(white)(m)("1 kPa = 0.00986 atm")$

According to the Ideal Gas equation,

$P V = n R T$ $PV = nRT$
$R = 0.08206 L . {atm.K}^{- 1} . {mol}^{- 1}$ $R = "0.08206 L"."atm.K"^(-1)."mol"^(-1)$
$n = {no. of moles of O}_{2}$ $n = "no. of moles of O"_2$

$P V = n R T$ $PV= nRT$

$n = \frac{P V}{R T}$ $n = (PV)/(RT)$

$n = \frac{1.48 atm \times 10.0 L}{0.08206 L\cdotatm \cdot K^{- 1} \cdot {mol}^{- 1} \times 423.15 K} = 0.4262 mol$ $n = ("1.48 atm" xx "10.0 L")/("0.08206 L·atm"·"K"^(-1)·"mol"^(-1) xx 423.15 K) = "0.4262 mol"$

$O_{2} = 0.4262 mol m m m m m m m m m m m m m m m m (2)$ $"O"_2 = "0.4262 mol"color(white)(mmmmmmmmmmmmmmmm)(2)$

According to the balanced equation

3 mol of $O_{2}$ $"O"_2$ is produced from the decomposition of 2 mol of ${KClO}_{3}$ $"KClO"_3$

0.4262 mol of $O_{2}$ $"O"_2$ will be produced from

$\frac{2 \times 0.4262}{3} l {mol KClO}_{3} = {0.2841 mol of KClO}_{3}$ $(2 xx 0.4262)/3 color(white)(l)"mol KClO"_3 = "0.2841 mol of KClO"_3$

Therefore ${KClO}_{3} = 0.2841 mol$ $"KClO"_3 = "0.2841 mol"$

${molar mass of KClO}_{3} = 122.55 g/mol$ $"molar mass of KClO"_3 = "122.55 g/mol"$

${mass of KClO}_{3} = 122.55 g/mol \times 0.2841 mol = 34.82 g$ $"mass of KClO"_3 = "122.55 g/mol" xx "0.2841 mol" = bb("34.82 g")$

Answer 2 · 2017-02-12T11:23:09

The reaction requires 34.8 g of ${KClO}_{3}$ $"KClO"_3$ .

Explanation:

There are four steps involved in this problem:

Write the balanced equation for the reaction.
Use the Ideal Gas Law to calculate the moles of $O_{2}$ $"O"_2$ .
Use the molar ratio of ${KClO}_{3} : O_{2}$ $"KClO"_3:"O"_2$ from the balanced equation to calculate the moles of ${KClO}_{3}$ $"KClO"_3$ .
Use the molar mass of ${KClO}_{3}$ $"KClO"_3$ to calculate the mass of ${KClO}_{3}$ $"KClO"_3$ .

Let's get started.

Step 1. Write the balanced chemical equation.

${2KClO}_{3} \to 2KCl + {3O}_{2}$ $"2KClO"_3 → "2KCl" + "3O"_2$

Step 2. Calculate the moles of $O_{2}$ $"O"_2$ .

The Ideal Gas Law is

$color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "$

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

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 $"KClO"_3$ .

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

4. Calculate the mass of $"KClO"_3$

$"Mass of KClO"_3 = 0.2842 color(red)(cancel(color(black)("mol KClO"_3))) × "122.55 g KClO"_3/(1 color(red)(cancel(color(black)("mol KClO"_3)))) ="34.8 g KClO"_3$

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Question #94034

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