<span>The theoretical yield for a reaction is calculated based on the limiting reagent. This allows researchers to determine how much product can actually be formed based on the reagents present at the beginning of the reaction.</span>
<span>The actual yield will never be 100 percent due to limitations.</span>
Answer:
The given reaction is a combustion reaction of benzene,
C
6
H
6
. From its balanced chemical equation,
2
C
6
H
6
+
15
O
2
→
12
C
O
2
+
6
H
2
O
,
the mass of carbon dioxide
(
C
O
2
)
produced from 20 grams (g) of
C
6
H
6
is determined through the molar mass of the two compounds, given by,
M
M
C
O
2
=
44.01
g
/
m
o
l
M
M
C
6
H
6
=
78.11
g
/
m
o
l
and their mole ratio:
12
m
o
l
C
O
2
2
m
o
l
C
6
H
6
→
6
m
o
l
C
O
2
1
m
o
l
C
6
H
6
With this,
m
a
s
s
o
f
C
O
2
=
(
20
g
C
6
H
6
)
(
1
m
o
l
C
6
H
6
78.11
g
C
6
H
6
)
(
6
m
o
l
C
O
2
1
m
o
l
C
6
H
6
)
(
44.01
g
C
O
2
1
m
o
l
C
O
2
)
=
(
20
)
(
6
)
(
44.01
)
g
C
O
2
78.11
=
5281.2
g
C
O
2
78.11
m
a
s
s
o
f
C
O
2
=
67.6
g
C
O
2
Therefore, the mass in grams of
C
O
2
formed from 20 grams of
C
6
H
6
is
67.6
g
C
O
2
.
it is a problem of app
Answer:
150
Explanation:
- C₄H₂OH + 6O2 → 4CO2 + 5H₂O
We can <u>find the equivalent number of O₂ molecules for 100 molecules of CO₂</u> using a <em>conversion factor containing the stoichiometric coefficients of the balanced reaction</em>, as follows:
- 100 molecules CO₂ *
= 150 molecules O₂
150 molecules of O₂ would produce 100 molecules of CO₂.
The answer is gonna be the last one :)
