Question

What is the weight of KClO3

required to get 16.8 Litres of Oxygen at STP?​

Answer 1

Explanation:

2KClO

3

Δ

3O

2

+2KCl

2mole3mole2mole

Number of moles of O

2

at S.T.P.=

22400

V(ml)

=

22400

448

=0.02mole

Mass of 1 mole of O

2

= 32 grams

Mass of 0.02 mole of O

2

=32× 0.02 = 0.64 grams

Mass of KClO

3

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

Weight of KClO

3

to produce 3 moles of O

2

= 245 grams

Weight of KClO

3

to produce 0.02 moles of O

2

=

3

245×0.02

= 1.63 grams

Mass of KCl produced = Mass of KClO

3

-Mass of O

2

produced =1.63−0.64=0.993 grams

Weight of O

2

produced = 0.64 grams

Weight of KClO

3

originally taken = 1.63 grams

Weight of KCl produced = 0.993 grams