What weight of AgCl will be precipitated by adding solution of 4.77g NaCl to a solution of 5.77g of AgNO3?
Answers
4.87
gram
A
g
C
l
will be precipitated.
Explanation:
The first thing to do is to write down the reaction that is described in the question. Since the reactant and one of the product are given, we can write the balanced reaction:
N
a
C
l
+
A
g
N
O
3
→
A
g
C
l
+
N
a
N
O
3
In this type of questions, it is nice to write down a kind of 'scheme' in which we place all the information provided so we can have a clear look what to do. But first, it is important to calculate the masses into moles, before we on accident calculate with the wrong numbers. To do this, we need to molar masses of the compounds. This can be calculated by adding up the masses of the atoms in the compounds. We obtain:
A
g
C
l
=
1
×
A
g
+
1
×
C
l
=
107.9
u
+
35.45
u
=
143.35
u
N
a
C
l
=
1
×
N
a
+
1
×
C
l
=
22.99
u
+
34.45
u
=
57.44
u
A
g
N
O
3
=
1
×
A
g
+
1
×
N
+
3
×
O
=
107.9
u
+
14.01
u
+
3
×
16.00
u
=
169.91
u
Now we calculate a number of moles with the formula below.
m
o
l
=
m
a
s
s
m
o
l
a
r
m
a
s
s
I only display the calculation for
N
a
C
l
here.
m
o
l
=
4.77
(
g
r
a
m
)
57.44
(
g
r
a
m
m
o
l
)
=
0.083
a
m
o
l
N
a
C
l
A
g
N
O
3
=
0.034
a
m
o
l
Now we have the moles, we write the molar ratio below the reaction and the moles at the correct places. We obtain something like this:
a
a
N
a
C
l
+
A
g
N
O
3
→
A
g
C
l
+
N
a
N
O
3
a
a
a
reaction
a
a
a
a
1
a
a
b
:
a
a
a
1
a
a
a
b
:
a
a
a
1
a
a
:
a
a
a
b
1
a
a
.
a
a
molar ratio
a
a
0.083
a
a
a
0.034
a
a
a
a
a
a
a
?
a
a
a
a
a
a
a
a
a
a
a
a
a
moles
We don't need to do anything with the
N
a
N
O
3
in this question.
We calculate the question mark
?
by applying the law of conservation of mass. This can be done by:
mol
A
g
C
l
=
m
o
l
A
g
N
O
3
×
m
o
l
a
r
r
a
t
i
o
A
g
C
l
m
o
l
a
r
r
a
t
i
o
A
g
N
O
3
mol
A
g
C
l
=
0.034
×
1
1
=
0.034
Now we calculate the mass of
A
g
C
l
that is created. The molar mass was
143.35
u
mass
A
g
C
l
=
m
o
l
×
m
o
l
a
r
m
a
s
s
mass
A
g
C
l
=
0.034
m
o
l
×
143.35
g
r
a
m
m
o
l
=
4.87
a
g
r
a
m