what amount of zn will be required to produce H2 by its action on dillute H2SO4 which will be completely react with oxygen produced by heating 30 gram of KCLo3??.
Answers
Answer:
the answer is given in the pictures above.
Explanation:
Explanation:
Start with the only known quantity in this question
m(KClO3)=30lg
n(KClO3)=30lg
⋅1lmol138.55lg=0.217lmol
The balanced equation of the decomposition of potassium perchlorate
2lKClO3(s)Δ−−→2lKCl(s)+3lO2(g)
suggests the stoichiometric relationship
2lmollKClO
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
lmollH2
of hydrogen. All these
H2
came from the reaction between
Zn
and dilute sulfuric acid
H2SO4
as seen in the following equation
1lZn(s)+H2SO4(aq)→1lH2(g)+ZnSO4(aq)
where for each mole of
H2(g)
produced,
1lmol of Zn
is consumed. That is:
n(Zn)=n(H2)=0.651lmol
Hence the mass of
Znm(Zn)=n(Zn)⋅M(Zn)=42.56lg