Step 1. Write the balanced chemical equation.
The balanced equation is
"2KClO"_3 → "2KCl" + "3O"_22KClO3→2KCl+3O2
Step 2. Strategy
The problem is to convert grams of "KClO"_3KClO3 to millilitres of "O"_2O2.
We can use the flow chart below to help us.
![Flow Chart]()
(Adapted from www.lsua.us)
The process is:
(a) Use the molar mass to convert mass of "KClO"_3KClO3 to moles of "KClO"_3KClO3.
(b) Use the molar ratio (from the balanced equation) to convert moles of "KClO"_3KClO3 to moles of "O"_2O2.
(e) Use the Ideal Gas Law to convert moles of "O"_2O2 to volume of "O"_2O2.
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"_2grams of KClO3molar massml−−−−−−−→moles of KClO3molar ratioml−−−−−−−→moles of O2Ideal Gas Lawml−−−−−−−−−→volume of O2
The Calculations
(a) Moles of "KClO"_3KClO3
0.300 color(red)(cancel(color(black)("g KClO"_3))) × ("1 mol KClO"_3)/( 122.55 color(red)(cancel(color(black)("g KClO"_3)))) = "0.002 448 mol KClO"_3
(b) Moles of "O"_2
0.002 448color(red)(cancel(color(black)("mol KClO"_3))) × ("3 mol O"_2)/(2 color(red)(cancel(color(black)("mol KClO"_3)))) = "0.003 672 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 = "0.003 672 mol"
R = "0.082 06 L·atm·K"^"-1""mol"^"-1"
T = "(25 + 273.15) K" = "298.15 K"
P = 755 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.9934 atm"
∴ V = ("0.003 672" color(red)(cancel(color(black)("mol"))) × "0.082 06"color(white)(l) "L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 273.15color(red)(cancel(color(black)( "K"))))/(0.9934 color(red)(cancel(color(black)("atm")))) = "0.0829 L" = " 82.9 mL"
The volume of "O"_2 produced is "82.9 mL".
