Volume of O2 : 168 L
Given
Reaction
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)
5 moles of KCIO3
Required
volume of O2
Solution
From the equation, mol O2 :
= 3/2 x moles KClO3
= 3/2 x 5 moles
= 7.5 moles
Assumed at STP( 1 mol = 22.4 L) :
= 7.5 x 22.4 L
= 168 L