Answer:
Elemental S reacts with O2 to form SO3 according to the reaction 2S+3O2→2SO3 Part B: What is the theoretical yield of SO3 produced by the quantities described in Part A? Express your answer numerically in grams.
Part A: 1.88x10^23 O2 molecules are needed to react with 6.67 g of S.
We address the equation...
S
(
s
)
+
3
2
O
2
(
g
)
→
S
O
3
(
g
)
Explanation:
The question specifies that we got
1.88
×
10
23
dioxygen molecules
...i.e. a molar quantity of...
1.88
×
10
23
⋅
molecules
6.022
×
10
23
⋅
molecules
⋅
m
o
l
−
1
=
0.312
⋅
m
o
l
...
But we gots with respect to sulfur,
6.67
⋅
g
32.06
⋅
g
⋅
m
o
l
−
1
=
0.208
⋅
m
o
l
...
And a bit of arithmetic later, we establish that we got stoichiometric quantities of dioxygen, and sulfur….in the reaction we produce a mass of ………..
0.208
⋅
m
o
l
×
80.07
⋅
g
⋅
m
o
l
−
1
=
16.65
⋅
g
.
Note that when
sulfur trioxide
is made industrially (and this a very important commodity chemical), sulfur is oxidized to
S
O
2
, and this is then oxidized up to
S
O
3
with some catalysis...
S
O
2
(
g
)
+
1
2
O
2
(
g
)
V
2
O
5
−−→
S
O
3
(
g
)
S
O
3
(
g
)
+
H
2
O
(
l
)
→
H
2
S
O
4
(
a
q
)
sulfuric acid
The industrial sulfur cycle must be a dirty, smelly, unfriendly process. The process is undoubtedly necessary to support our civilization....
