1. For the balanced equation shown below, if the reaction of 40.8 grams of C6H6O3 produces a 39.0% yield, how many grams of H2O would be produced?
C6H6O3 + 6O2 6CO2 + 3H2O
Answers
Starting from the balanced chemical equation
C
6
H
6
O
3
+
6
O
2
→
6
C
O
2
+
3
H
2
O
we can see that we have a
1
:
3
mole ratio between
C
6
H
6
O
3
and
H
2
O
; that is, for every mole of
C
6
H
6
O
3
used in the reaction,
3
moles of
H
2
O
are produced.
SInce the reaction's percent yield is
39.0
%
, we can say that not all
C
6
H
6
O
3
was used in the reaction <=>
O
2
is the limiting reagent.
The number of
C
6
H
6
O
3
moles, knowing its molar mass is
126
g
m
o
l
, is
n
C
6
H
6
O
3
=
m
C
6
H
6
O
3
m
o
l
a
r
m
a
s
s
=
40.8
g
126
g
m
o
l
=
0.32
moles
Now, if all the
C
6
H
6
O
3
would have reacted, the number of moles and, subsequently, the mass of
H
2
O
produced would have been
n
H
2
O
=
3
⋅
n
C
6
H
6
O
3
=
3
⋅
0.32
=
0.96
moles, and
m
H
2
O
=
n
H
2
O
⋅
(
m
o
l
a
r
m
a
s
s
)
=
0.96
m
o
l
e
s
⋅
18
g
m
o
l
=
17.3
g
However, we know that the actual number of moles of
H
2
O
produced is
%
y
i
e
l
d
=
n
a
c
t
u
a
l
n
t
h
e
o
r
e
t
i
c
⋅
(
100
%
)
→
n
a
c
t
u
a
l
=
%
y
i
e
l
d
⋅
n
t
h
e
o
r
e
t
i
c
100
%
=
39.0
⋅
0.96
100
%
=
0.37
Therefore, the mass of
H
2
O
produced is
m
H
2
O
=
n
a
c
t
u
a
l
⋅
(
m
o
l
a
r
m
a
s
s
)
=
0.37
m
o
l
e
s
⋅
18
g
m
o
l
=
6.7
g