Calculate the mass of KClO3 necessary to produce 1.23 g of O2.
Answers
Answer:
2KClO
3
→2KCl+3O
2
Here if the volume of O
2
at STP is 1.12 L, then at STP 1 mol gas will have volume 22.4 L, so 1.12 L must
have
22.4
1
×1.12=
22.40
1.12
=
2240
112
=
20
1
moles.
It can be seen from the reaction that for 3 mole O
2
→2 moles of KCl produced.
So if 1 mole O
2
is produced →
3
2
mole of KCl must be produced along with.
∴ If O
2
yield is
20
1
mole during the reaction, then KCl must be
3
2
×
20
1
moles.
Hence KCl moles =
60
2
moles
As Molar mass of KCl=39+35.5=74.5 g.
so
30
2
moles of KCl must have weight =2/60×74.5=2.48 g.
Answered By
enistien
How satisfied are you with the answer?
This will help us to improve better
answr
Answer:
We first need our balanced equation:
KClO3→2KCl+3O.
Equation tells us, 3 moles of O2 are produced by 2 moles of KClO3 that is
- 3 moles of O2 →16×6=96g
- 2 moles of KClO3→245g
so,1.23 g are produced by 245/(96×1.23)=0.48g