Question

2 KClO2 (s) 0 2 KCl (s) + 3 02 (g)

Answer 1

Volume of O2 : 168 L

Further explanation

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