What volume of #O_2#, measured 82.1°C and 713 mm Hg, will be produced by the decomposition of 5.71 g of #KClO_3#?

1 Answer
Nov 15, 2016

Answer:

The reaction produces #"2.18 L of O"_2#.

Explanation:

Step 1. Write the balanced chemical equation.

The balanced equation is

#"2KClO"_3 → "2KCl" + "3O"_2#

Step 2. The Procedure

The problem is to convert grams of #"KClO"_3# to moles of #"O"_2# and volume of #"O"_2#.

The procedure is:

(a) Use the molar mass to convert the mass of #"KClO"_3# to moles of #"KClO"_3#.

(b) Use the molar ratio (from the balanced equation) to convert moles of #"KClO"_3# to moles of #"O"_2#.

(e) Use the Ideal Gas Law to convert moles of #"O"_2# to volume of #"O"_2#.

In equation form,

#"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 #"KClO"_3#

#5.71 color(red)(cancel(color(black)("g KClO"_3))) × ("1 mol KClO"_3)/(122.55 color(red)(cancel(color(black)("g KClO"_3)))) = "0.046 59 mol KClO"_3 #

(b) Moles of #"O"_2#

#"0.046 59"color(red)(cancel(color(black)("mol KClO"_3))) × ("3 mol O"_2)/(2 color(red)(cancel(color(black)("mol KClO"_3)))) = "0.069 89 mol O"_2#

(c) Volume of #"O"_2#

The Ideal Gas Law is

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

We can rearrange this to give

#V = (nRT)/P#

#n = "0.069 89 mol O"_2#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "(82.1 + 273.15) K" = "355.25 K"#
#P = 711 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.9355 atm"#

#V = ("0.069 89" color(red)(cancel(color(black)("mol"))) × "0.082 06 L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 355.25color(red)(cancel(color(black)( "K"))))/(0.9355 color(red)(cancel(color(black)("atm")))) = "2.18 L"#

The volume of #"O"_2# produced is #"2.18 L"#.