Question

Question #94034

Answer 1

The mass of potassium chlorate = 34.8 g

Explanation:

`"2KClO"_3 rarr "2KCl" + "3O"_2 color(white)(mmmmmmmmmmmm)(1)`

`"Volume O"_2 = "10.0 L"`
`T = "(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,

`PV = nRT`
`R = "0.08206 L"."atm.K"^(-1)."mol"^(-1)`
`n = "no. of moles of O"_2`

`PV= nRT`

`n = (PV)/(RT)`

`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"color(white)(mmmmmmmmmmmmmmmm)(2)`

According to the balanced equation

3 mol of `"O"_2` is produced from the decomposition of 2 mol of `"KClO"_3`

0.4262 mol of `"O"_2` will be produced from

`(2 xx 0.4262)/3 color(white)(l)"mol KClO"_3 = "0.2841 mol of KClO"_3`

Therefore `"KClO"_3 = "0.2841 mol"`

`"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")`

Answer 2

The reaction requires 34.8 g of `"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 `"O"_2`.
3. Use the molar ratio of `"KClO"_3:"O"_2` from the balanced equation to calculate the moles of `"KClO"_3`.
4. Use the molar mass of `"KClO"_3` to calculate the mass of `"KClO"_3`.

Let's get started.

Step 1. Write the balanced chemical equation.

`"2KClO"_3 → "2KCl" + "3O"_2`

Step 2. Calculate the moles of `"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`