17. 100 mL of 0.1 M NaCl solution is mixed with 100 mL of 0.2 M AgNO, solution. The
reaction is as follows:
NaCl(aq) + AgNO3(aq)
→ NaNO3(aq) + AgCl(s)
(a)
Identify the limiting reagent.
(b)
Find out the mass of AgCl precipitate formed. (Molar mass of AgCl = 143.5 g
mol-1)
1+1=2
Answers
Answer:
In the reaction shown above, if we mixed 123 mL of a 1.00 M solution of NaCl with 72.5 mL of a 2.71 M solution of AgNO 3 , we could calculate the moles (and hence, the mass) of AgCl that will be formed as follows: To determine the concentration of potassium nitrate in the final solution, we need to note that two moles of potassium nitrate are formed for every mole of PbI 2 , or a stoichiometric ratio of \[\left ( \frac{2\: moles\: KNO_{3}}{1\: mole\: PbI_{2}} \right )\]
Explanation:
Answer:
17.6 g AgCl(s) is formed in the reaction.
Explanation:
First, we need to write out our balanced reaction equation:
A
g
N
O
3
(
a
q
)
+
N
a
C
l
(
a
q
)
→
A
g
C
l
(
s
)
+
N
a
N
O
3
(
a
q
)
The next step, as in any calculation involving stoichiometry, is to determine our limiting reactant. We can do this by converting both of our reactants into moles:
123 mL
N
a
C
l
×
1 L
1000 mL
×
1.00 mol
N
a
C
l
1 L
=
0.123 mol
N
a
C
l
72.5 mL
A
g
N
O
3
×
1 L
1000 mL
×
2.71 mol
A
g
N
O
3
1 L
=
0.196 mol
A
g
N
O
3
We can see from our reaction equation that AgNO3 and NaCl react in a 1:1 ratio. Because there are fewer moles of NaCl present in solution, NaCl is our limiting reactant. We can now solve for the mass of AgCl formed:
123 mL
N
a
C
l
×
1 L
1000 mL
×
1.00 mol
N
a
C
l
1 L
×
1 mol
A
g
C
l
1 mol
N
a
C
l
×
143 g
1 mol
A
g
C
l
=
17.6 g
A
g
C
l
,
Therefore, 17.6 g AgCl(s) is formed in the reaction.
To sum up: we converted to each reactant’s moles by using the given concentrations as conversion factors, expressing molarity as mol/L; once we found our limiting reactant, we converted through to grams of AgCl formed.