what is the percent yield of CuS if the given reaction starts with 15.5 grams of Na2SW and 12.1 grams of CuSO? the actual amount of CuS produced is 3.05 grams
Answers
Answer:
N
a
2
S
+
C
u
S
O
4
→
N
a
2
S
O
4
+
C
u
S
Given data
M
a
s
s
o
f
N
a
2
S
=
15.5
g
M
a
s
s
o
f
C
u
S
O
4
=
12.1
g
A
c
t
u
a
l
a
m
o
u
n
t
o
f
C
u
S
=
3
.05
g
Calculation
No. of moles of
N
a
2
S
:
n
=
M
a
s
s
o
f
N
a
2
S
M
o
l
a
r
m
a
s
s
o
f
N
a
2
S
=
15
.5
g
78
.0452
g
/
m
o
l
=
0
.198
m
o
l
No. of moles of
C
u
S
O
4
:
n
=
M
a
s
s
o
f
C
u
S
O
4
M
o
l
a
r
m
a
s
s
o
f
C
u
S
O
4
=
12.1
g
159.609
g
/
m
o
l
=
0.075
m
o
l
Here, in the reaction the number of moles of
C
u
S
O
4
is less than the moles of
N
a
2
S
. Hence,
C
u
S
O
4
is the limiting reagent in the reaction.
In the reaction, one mole of
N
a
2
S
react with one mole of
C
u
S
O
4
to form one mole
C
u
S
. Therefore, 0.075 mol of
C
u
S
O
4
react with
N
a
2
S
to form 0.075 mol of
C
u
S
.
Theoretical amount of
C
u
S
produced in the reaction:
M
a
s
s
=
n
×
M
o
l
a
r
m
a
s
s
=
0
.075
×
95
.611
g
/
m
o
l
=
7
.171
g
Percent yield of the reaction:
P
e
r
c
e
n
t
Y
i
e
l
d
=
A
c
t
u
a
l
y
i
e
l
d
T
h
e
o
r
e
t
i
c
a
l
y
i
e
l
d
×
100
%
=
3.05
g
7.171
g
×
100
%
=
42.53
%
Conclusion
Hence, the limiting reagent is
C
u
S
O
4
in the reaction and percent yield is
42.53
%
.