what amount of zn will be required to produced h2 by its action on dil h2so4 which will completely react with the oxygen , produced by heating 30g of kclo3
Answers
Answer:
Start with the only known quantity in this question
m
(
KClO
3
)
=
30
l
g
n
(
KClO
3
)
=
30
l
g
⋅
1
l
mol
138.55
l
g
=
0.217
l
mol
The balanced equation of the decomposition of potassium perchlorate
2
l
KClO
3
(
s
)
Δ
−−→
2
l
KCl
(
s
)
+
3
l
O
2
(
g
)
suggests the stoichiometric relationship
2
l
mol
l
KClO
3
decomposes to produce
3
l
mol
l
O
2
(
g
)
Hence the amount of oxygen required for the complete combustion of the unknown amount of
H
2
(
g
)
produced would be
n
(
O
2
)
=
0.217
l
mol
l
KClO
3
⋅
3
l
mol
l
O
2
2
l
mol
l
KClO
3
=
0.326
l
mol
l
O
2
Oxygen reacts with hydrogen by the equation
1
l
O
2
(
g
)
+
2
l
H
2
(
g
)
*
−−→
2
l
H
2
O
(
g
)
at a
1
:
2
ratio, meaning that the combustion would consume
n
(
H
2
)
=
0.326
l
mol
l
O
2
⋅
2
l
H
2
1
l
O
2
=
0.651
l
mol
l
H
2
of hydrogen. All these
H
2
came from the reaction between
Zn
and dilute sulfuric acid
H
2
SO
4
as seen in the following equation
1
l
Zn
(
s
)
+
H
2
SO
4
(
a
q
)
→
1
l
H
2
(
g
)
+
ZnSO
4
(
a
q
)
where for each mole of
H
2
(
g
)
produced,
1
l
mol
of
Zn
is consumed. That is:
n
(
Zn
)
=
n
(
H
2
)
=
0.651
l
mol
Hence the mass of
Zn
m
(
Zn
)
=
n
(
Zn
)
⋅
M
(
Zn
)
=
42.56
l
g