Question #94034

2 Answers

Answer:

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")#

Feb 12, 2017

Answer:

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#