how many moles of O2 are produced when 6.0 mol of KCIO3 decompose completely
kuch to h jo tu keh do_____
complete the linees
Answers
Answer:
9.0
l
mol
Explanation:
Potassium chlorate
KClO
3
decomposes to produce potassium chloride
KCl
and oxygen
O
2
. Balancing the equation
KClO
3
→
KCl
+
O
2
(not balanced)
based on the fact that the number of moles of oxygen atoms should be the same on both side of the equation will give
KClO
3
→
KCl
+
3
2
O
2
2
KClO
3
→
2
KCl
+
3
O
2
Hence the ratio of number of moles of particles
n
(
O
2
)
n
(
KClO
3
)
=
3
2
n
(
O
2
)
=
n
(
KClO
3
)
⋅
n
(
O
2
)
n
(
KClO
3
)
=
3
2
⋅
n
(
KClO
3
)
Therefore the decomposition of
6.0
l
mol
of
KClO
3
would yield
6.0
⋅
n
(
O
2
)
n
(
KClO
3
)
=
6.0
⋅
3
2
=
9.0
l
mol
Note that the species
KCl
is highly stable. Both the potassium cation
K
+
and the chloride anion
Cl
−
have attained a noble gas octet configuration of valence shell electrons. Chemical reactions tend to favor the most chemically stable combination. Therefore this decomposition reaction would eventually produce
KCl
rather than other salts that contain oxygen.
Answer:
Always start with a balanced equation.
2KClO3 → 2KCl + 3O2
Use stoichiometry to calculate the moles of O2 produced when 6.4 moles of KClO3 decompose.
Multiply 6.4 moles of KClO3 by the mole ratio between KClO3 and O2 in the balanced equation, so that moles KClO3 cancel, leaving moles O2.
6.4 mol KClO3 × 3 mol O2/2 mol KClO3 = 9.6 mol O2
When 6.4 moles of KClO3 decompose, 9.6 moles of O2 are produced.