**Step 1. Write the balanced chemical equation.**

The balanced equation is

#"2KClO"_3 → "2KCl" + "3O"_2#

**Step 2. Strategy**

The problem is to convert grams of #"KClO"_3# to millilitres of #"O"_2#.

We can use the flow chart below to help us.

(Adapted from www.lsua.us)

The process is:

**(a)** Use the **molar mass** to convert 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#**

#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"#.