How much amount of sulphur can be obtained by the reaction of 1 mol of so2 with 22.4 l of h2s at stp?
Answers
Answer:
Explanation:
The sulfide in
hydrogen sulfide
is OXIDIZED
(
i
)
to elemental sulfur, and the
sulfur dioxide
is REDUCED to elemental sulfur......
(
i
i
)
S
2
−
→
S
+
2
e
−
(
i
)
S
O
2
+
4
H
+
+
4
e
−
→
S
+
2
H
2
O
(
i
i
)
Both mass and charge are balanced in each reaction, as indeed they must be if we purport to represent chemical reality.....and so we simply add
2
×
(
i
)
+
(
i
i
)
to eliminate the electrons to give........
2
S
2
−
+
S
O
2
+
4
H
+
+
4
e
−
→
3
S
+
2
H
2
O
+
4
e
−
And we could simplify further to give.......
2
H
2
S
+
S
O
2
→
3
S
+
2
H
2
O
And now (finally) we can address your question.
NTP
specifies a molar volume of
24.06
⋅
L
⋅
m
o
l
−
1
......
With respect to
S
O
2
we have a molar quantity of
11.2
⋅
L
24.06
⋅
L
⋅
m
o
l
−
1
=
0.466
⋅
m
o
l
...
.
. And thus
S
O
2
is the limiting reagent and
hydrogen sulfide
is in excess.
And so we gets
3
×
an equivalent quantity of
S
, i.e.
3
⋅
S
1
⋅
S
O
2
×
0.466
⋅
m
o
l
s
×
32.06
⋅
g
⋅
m
o
l
−
1
=
44.8
⋅
g
Hopenit helps and marks a s brainlest answer.....
So2 + 2H2S ---------> 3S + 2H 2O
1 mol 22.4 L
22.4/ 22.4 = 1 mol
1 mol 1 mol
1 mol H2S reaction
1/2 mol O2 3/2 mol formed
3/2 * 32 = 48g
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