Question

In following reaction ,H= k87J is used up. Find if oxygen liberated is 32g ,than KCL produced is ------g.

2KClO3=2KCl+3O2​

Answer 1

Solution:

2KClO3Δ3O2+2KCl2mole3mole2mole

Number of moles of O2 at S.T.P.=22400V(ml)=22400448=0.02mole

Mass of 1 mole of O2 = 32 grams

Mass of 0.02 mole of O2=32× 0.02 = 0.64 grams

Mass of KClO3=moles(n)×molecular weight=2× 122.5 = 245 grams

Weight of KClO3 to produce 3 moles of O2 = 245 grams

Weight of KClO3 to produce 0.02 moles of O2=3245×0.02= 1.63 grams

Mass of KCl produced = Mass of KClO3-Mass of O2 produced =1.63−0.64=0.993 grams

Weight of O2 produced = 0.64 grams

Weight of KClO3 originally taken = 1.63 grams

Weight of KCl produced = 0.993 grams