KClO3 on heating decomposes to give KCl and Oxygen. What is the volume of oxygen liberated at NTP by 0.1mol of KClO3?
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Solution
KClO3 -------> KCl + 3/2 O2
1 mol KClO3 produces 3/2 mol O2 0.1 mol KClO3 will produce 3/2*0.1 = 0.15 mol O2 At STP , 1 mol O2 has volume = 22.4L 0.15 mol O2 will have volume = 0.15*22.4 m= 3.360 L Answer: volume of O2 produced = 3.360L