CHEM 122 - GENERAL CHEMISTRY
II - Precipitation calculations
Question Bank - Set 5
Liberty University
Question 1
Question
Calculate the mass of silver chloride (AgCl) that can be formed when 50.0 mL
of 0.200 M silver nitrate (AgNO3) solution is mixed with 75.0 mL of 0.150 M
sodium chloride (NaCl) solution. Assume the reaction goes to completion and
that silver chloride is insoluble in water.
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium chloride to form silver chloride:
AgNO3+ NaCl →AgCl + NaNO3
Step 2: Determine the limiting reactant by calculating the moles of each
reactant:
For silver nitrate:
moles of AgNO3= volume (L) ×molarity = 0.050 L ×0.200 mol/L = 0.010 mol
For sodium chloride:
moles of NaCl = volume (L) ×molarity = 0.075 L ×0.150 mol/L = 0.01125 mol
Since there are more moles of NaCl than AgNO3, AgNO3is the limiting
reactant.
Step 3: Calculate the mass of silver chloride formed using the mole ratio
from the balanced chemical equation:
mol of AgCl formed = mol of AgNO3= 0.010 mol
mass of AgCl = mol of AgCl ×molar mass of AgCl
The molar mass of AgCl is the sum of the atomic masses of silver (Ag) and
chlorine (Cl):
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Therefore,
mass of AgCl = 0.010 mol ×143.32 g/mol = 1.4332 g
The mass of silver chloride that can be formed when 50.0 mL of 0.200 M
silver nitrate solution is mixed with 75.0 mL of 0.150 M sodium chloride solution
is 1.4332 g.
Question 2
Question
A certain chemical reaction produces a precipitate when two solutions are mixed.
The balanced chemical equation is given by:
BaCl2(aq) + Na2SO4(aq)→BaSO4(s) + 2NaCl(aq)
If 500.0 mL of a 0.200 M solution of BaCl2 is mixed with 300.0 mL of a
0.150 M solution of Na2SO4, what mass of BaSO4 will be produced? (Assume
that all of the BaCl2 and Na2SO4 react to produce BaSO4.)
Solution
Step 1: Calculate the moles of BaCl2 and Na2SO4 in the solutions. Given:
Volume of BaCl2solution = 500.0 mL = 0.5000 L
Molarity of BaCl2solution = 0.200 M
Volume of Na2SO4solution = 300.0 mL = 0.3000 L
Molarity of Na2SO4solution = 0.150 M
Moles of BaCl2:
Moles of BaCl2= Molarity ×Volume in Liters
Moles of BaCl2= 0.200 mol/L ×0.5000 L
Moles of BaCl2= 0.1000 mol
2
Moles of Na2SO4:
Moles of Na2SO4= Molarity ×Volume in Liters
Moles of Na2SO4= 0.150 mol/L ×0.3000 L
Moles of Na2SO4= 0.0450 mol
Step 2: Determine the limiting reactant. From the balanced chemical equa-
tion, we can see that 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to produce
1 mole of BaSO4. Since the mole ratio is 1:1, the limiting reactant will be the
one that is completely consumed first in the reaction.
Initially, we have: - Moles of BaCl2: 0.1000 mol - Moles of Na2SO4: 0.0450
mol
To find the limiting reactant, we compare the moles of product that each
reactant could produce. Since the mole ratio between BaCl2 and Na2SO4 is
1:1: 1. For BaCl2: Moles of BaSO4 produced = 0.1000 mol 2. For Na2SO4:
Moles of BaSO4 produced = 0.0450 mol
Since Na2SO4 produces fewer moles of BaSO4, Na2SO4 is the limiting re-
actant.
Step 3: Calculate the mass of BaSO4 produced. Given:
Molar mass of BaSO4= 137.3 g/mol
Moles of BaSO4 produced will be the same as moles of Na2SO4 used in the
reaction:
Moles of BaSO4= 0.0450 mol
Finally, we calculate the mass of BaSO4 produced:
Mass of BaSO4= Moles of BaSO4×Molar mass of BaSO4
Mass of BaSO4= 0.0450 mol ×137.3 g/mol
Mass of BaSO4= 6.18 g
Therefore, the mass of BaSO4 produced when 500.0 mL of a 0.200 M solution
of BaCl2 is mixed with 300.
Question 3
Question
A solution is prepared by mixing 150 mL of a 0.2 M lead(II) nitrate solution with
200 mL of a 0.3 M potassium iodide solution. Calculate the mass of lead(II)
iodide that precipitates. (Assume the reaction goes to completion)
3
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and potassium iodide. The balanced chemical equation for the reaction
is:
Pb(NO3)2(aq)+ 2KI(aq)→PbI2(s)+ 2KNO3
Step 2: Determine the limiting reactant. First, calculate the number of
moles of each reactant: For lead(II) nitrate (Pb(NO3)2): Number of moles =
Concentration (M) ×Volume (L) Number of moles = 0.2 mol/L ×0.150 L =
0.03 moles For potassium iodide (KI): Number of moles = Concentration (M)
×Volume (L) Number of moles = 0.3 mol/L ×0.200 L = 0.06 moles
Since we need 2 moles of KI for every mole of Pb(NO3)2, the limiting reactant
is Pb(NO3)2.
Step 3: Calculate the mass of lead(II) iodide precipitated. From the balanced
chemical equation, we see that 1 mole of Pb(NO3)2 produces 1 mole of PbI2.
The molar mass of PbI2 is 461 g/mol. Number of moles of PbI2 formed =
Number of moles of Pb(NO3)2 Number of moles of PbI2 = 0.03 moles Mass of
PbI2 formed = Number of moles of PbI2 ×Molar mass of PbI2 Mass of PbI2
formed = 0.03 moles ×461 g/mol = 13.83 g
Therefore, the mass of lead(II) iodide that precipitates is 13.83 grams.
Question 4
Question
Calculate the mass of barium sulfate (BaSO4) that precipitates when 50.0 mL
of 0.200 M barium chloride (BaCl2) reacts with excess sodium sulfate (Na2SO4)
according to the following balanced chemical equation:
BaCl2+ Na2SO4−→ BaSO4+ 2NaCl
Solution
Step 1: Write the balanced chemical equation and identify the limiting reactant.
The balanced chemical equation is given as:
BaCl2+ Na2SO4−→ BaSO4+ 2NaCl
This equation indicates that 1 mole of BaCl2reacts with 1 mole of Na2SO4to
produce 1 mole of BaSO4.
Given that the volume of BaCl2is 50.0 mL and the concentration is 0.200
M, we need to calculate the number of moles of BaCl2:
moles of BaCl2= concentration×volume = 0.200 mol/L×0.0500 L = 0.0100 mol
Since the ratio of BaCl2to Na2SO4is 1:1, the number of moles of Na2SO4
required to react with all the BaCl2is also 0.0100 mol.
4
Step 2: Calculate the mass of BaSO4precipitated. From the balanced chem-
ical equation, we see that 1 mole of BaSO4is produced for every 1 mole of BaCl2
reacted.
The molar mass of BaSO4is calculated as:
Ba = 137.33 g/mol
S = 32.07 g/mol
4×O = 4 ×16.00 g/mol = 64.00 g/mol
Adding these up, we get:
137.33 + 32.07 + 64.00 = 233.40 g/mol
The mass of BaSO4that precipitates can be calculated as:
mass = moles ×molar mass = 0.0100 mol ×233.40 g/mol = 2.33 g
Therefore, the mass of barium sulfate (BaSO4) that precipitates is 2.33 g.
Question 5
Question
Calculate the molarity of MgCl2in a solution if 3.50 grams of MgCl2is dissolved
in enough water to make 250.0 mL of solution.
(Given: molar mass of MgCl2= 95.21 g/mol)
Solution
Step 1: Calculate the number of moles of MgCl2.
moles of MgCl2=mass
molar mass =3.50 g
95.21 g/mol
Step 2: Convert the solution volume to liters.
Volume of solution = 250.0 mL ×1 L
1000 mL
Step 3: Calculate the molarity of MgCl2.
Molarity = moles
volume in liters =3.50 g/95.21 g/mol
250.0 mL/1000 mL/L
5
Question 6
Question
A solution is prepared by dissolving 25.0 g of silver nitrate (AgNO3) in enough
water to make 0.250 L of solution. To this, excess sodium chloride (NaCl) is
added resulting in the precipitation of silver chloride (AgCl). If the mass of the
precipitate formed is 15.0 g, what is the concentration of silver ions (Ag+) in
the solution expressed in mol/L? (Assume complete precipitation of silver as
silver chloride)
(Hint: The balanced chemical equation for the precipitation of AgCl is
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq))
Solution
Step 1: Calculate the moles of AgNO3in the solution. Given mass of AgNO3:
25.0 g Molar mass of AgNO3: 107.87 g/mol + 14.01 g/mol + 3(16.00 g/mol) =
169.87 g/mol
Number of moles of AgNO3=25.0 g
169.87 g/mol = 0.1471 mol
Step 2: Determine the moles of Ag in the solution. From the balanced
equation, 1 mole of AgNO3produces 1 mole of Ag ions. Therefore, moles of
Ag+= 0.1471 mol
Step 3: Calculate the concentration of Ag+ions. Volume of solution = 0.250
L Concentration of Ag+=moles of Ag+
volume of solution =0.1471 mol
0.250L= 0.5884 mol/L
Therefore, the concentration of silver ions (Ag+) in the solution is 0.5884
mol/L.
Question 7
Question
A solution contains 0.1 M barium chloride and 0.2 M sodium sulfate. Calculate
the concentration of barium ion in the solution after precipitation of barium
sulfate.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between barium chloride and sodium sulfate.
BaCl2+ Na2SO4→BaSO4(s) + 2NaCl
Step 2: Determine the limiting reactant by comparing the number of moles
of BaCl2and Na2SO4.
Step 3: Calculate the number of moles of BaSO4formed using the stoichiom-
etry of the balanced equation.
6
Step 4: Since barium sulfate is insoluble, it will precipitate. Calculate the
concentration of Ba2+ ions in the solution after precipitation.
Step 5: Using the initial volume of the solution, calculate the final concen-
tration of Ba2+ ions.
Step 6: Express the concentration of Ba2+ ions in the solution.
Question 8
Question
A solution contains 25.0 g of potassium iodide (KI) in 150.0 mL of water. If
silver nitrate (AgNO3) is added to the solution, how many grams of silver iodide
(AgI) will precipitate out? Assume that the reaction goes to completion and
that only the formation of AgI is considered.
Solution
Step 1: Write the balanced chemical equation for the reaction between KI and
AgNO3to form AgI:
KI + AgNO3→AgI + KNO3
Step 2: Calculate the number of moles of KI in the solution:
Number of moles of KI = Mass (g)
Molar mass (g/mol) =25.0 g
166.00 g/mol = 0.1506 mol
Step 3: Find the limiting reagent by considering the stoichiometry of the
reaction. Since the reaction proceeds to completion, the limiting reagent is the
one that produces the least amount of AgI. Calculate the number of moles of
AgNO3needed to react with all of the KI:
Number of moles of AgNO3=Number of moles of KI
1= 0.1506 mol
Step 4: Calculate the mass of AgI produced:
Mass of AgI = Number of moles of AgI×Molar mass of AgI = 0.1506 mol×234.77 g/mol = 35.39 g
Therefore, 35.39 grams of silver iodide (AgI) will precipitate out.
Question 9
Question
A chemist is conducting a precipitation reaction in which 50.0 mL of 0.200 M
silver nitrate (AgNO3) is mixed with 75.0 mL of 0.150 M sodium chloride (NaCl).
Calculate the maximum mass of silver chloride (AgCl) that could precipitate.
(Given: molar mass of AgCl = 143.32 g/mol)
7
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between silver nitrate and sodium chloride:
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reactant by calculating the moles of each
reactant:
Moles of AgNO3:
Moles = Volume ×Molarity = 0.0500 L ×0.200 mol/L = 0.0100 mol
Moles of NaCl:
Moles = Volume ×Molarity = 0.0750 L ×0.150 mol/L = 0.0113 mol
Step 3: Use stoichiometry to determine the moles of AgCl that can be
formed: Since the reaction is 1:1 between AgNO3and AgCl, the moles of AgCl
formed will be 0.0100 mol (from AgNO3).
Step 4: Calculate the mass of AgCl formed using its molar mass:
Mass = Moles ×Molar mass = 0.0100 mol ×143.32 g/mol = 1.43 g
Therefore, the maximum mass of AgCl that could precipitate is 1.43 g.
Question 10
Question
A solution was prepared by mixing 50.0 mL of 0.200 M calcium chloride (CaCl2)
with 75.0 mL of 0.100 M sodium carbonate (Na2CO3). Calculate the mass of
the precipitate formed, assuming the reaction goes to completion.
Solution
Step 1: Write the balanced chemical equation for the reaction between calcium
chloride and sodium carbonate:
CaCl2+ Na2CO3→CaCO3+ 2NaCl
Step 2: Determine the limiting reagent by calculating the moles of each
reactant. For calcium chloride (CaCl2):
moles = molarity ×volume (L) = 0.200 mol/L ×0.0500 L = 0.0100 mol
For sodium carbonate (Na2CO3):
moles = molarity ×volume (L) = 0.100 mol/L ×0.0750 L = 0.00750 mol
8
Since 1 mole of CaCl2reacts with 1 mole of Na2CO3, Na2CO3is the limiting
reagent with 0.00750 mol.
Step 3: Calculate the theoretical yield of calcium carbonate (CaCO3) using
the limiting reagent:
moles of CaCO3= 0.00750 mol
Step 4: Convert the moles of calcium carbonate to grams using its molar
mass. The molar mass of CaCO3is:
1×Ca + 1 ×C+3×O = 40.08 + 12.01 + 3 ×16.00 = 100.09 g/mol
Therefore, the mass of CaCO3formed is:
mass = moles ×molar mass = 0.00750 mol ×100.09 g/mol = 0.751 g
Thus, the mass of the precipitate formed, calcium carbonate, is 0.751 g.
Question 11
Question
Calculate the mass of barium sulfate (BaSO4) that can be formed when 250.0
mL of a 0.200 M solution of barium chloride (BaCl2) is mixed with 300.0 mL
of a 0.150 M solution of sodium sulfate (Na2SO4). Assume the reaction goes to
completion.
Solution
Step 1: Write the balanced chemical equation for the reaction between barium
chloride and sodium sulfate to form barium sulfate and sodium chloride:
BaCl2+ Na2SO4→BaSO4+ 2NaCl
Step 2: Determine the limiting reactant by calculating the moles of each
reactant. The moles of barium chloride (BaCl2) is calculated as:
moles of BaCl2= Molarity ×Volume (L)
moles of BaCl2= 0.200 mol/L ×0.250 L = 0.0500 mol
The moles of sodium sulfate (Na2SO4) is calculated as:
moles of Na2SO4= Molarity ×Volume (L)
moles of Na2SO4= 0.150 mol/L ×0.300 L = 0.0450 mol
Since the stoichiometry of the reaction is 1:1 between BaCl2and Na2SO4,
Na2SO4is the limiting reactant.
9
Step 3: Calculate the theoretical yield of barium sulfate using the mole ratio
between Na2SO4and BaSO4.
moles of BaSO4= 0.0450 mol ×1 mol BaSO4
1 mol Na2SO4
= 0.0450 mol
Step 4: Calculate the mass of barium sulfate formed using its molar mass
(233.39 g/mol).
mass of BaSO4= 0.0450 mol ×233.39 g/mol = 10.50 g
Therefore, the mass of barium sulfate that can be formed is 10.50 g.
Question 12
Question
Calculate the concentration of chloride ions in a solution prepared by mixing
50.0 mL of a 0.200 M calcium chloride (CaCl2) solution with 100.0 mL of a
0.100 M sodium chloride (NaCl) solution. Assume the volumes are additive and
no volume changes occur upon mixing.
Solution
Step 1: Calculate the moles of chloride ions from each solute solution using the
formula n=C×V, where nis the number of moles, Cis the concentration,
and Vis the volume in liters.
For CaCl2:nCaCl2= 0.200 M ×0.0500 L = 0.010 mol
For NaCl: nNaCl = 0.100 M ×0.1000 L = 0.010 mol
Step 2: Determine the total moles of chloride ions in the solution by adding
the moles from both solutes. Total moles of chloride ions = nCaCl2+nNaCl =
0.010 mol + 0.010 mol = 0.020 mol
Step 3: Calculate the total volume of the solution. Vtotal = 0.0500 L +
0.1000 L = 0.1500 L
Step 4: Calculate the final concentration of chloride ions using the formula
Cfinal =ntotal
Vtotal .Cfinal =0.020 mol
0.1500 L = 0.133 M
Therefore, the concentration of chloride ions in the final solution is 0.133 M.
Question 13
Question
Calculate the solubility product constant (Ksp) of lead(II) chloride from the
solubility of lead(II) chloride in water at 25◦C, which is 4.53 x 10−3M.
10
Solution
Step 1: Write the balanced equation for the dissolution of lead(II) chloride in
water.
PbCl2(s)⇌Pb2+(aq) + 2Cl−(aq)
Step 2: Write the expression for the solubility product constant (Ksp).
Ksp = [Pb2+][Cl−]2
Step 3: List the known values from the question.
[Pb2+]=4.53 ×10−3M
[Cl−]=2×4.53 ×10−3= 9.06 ×10−3M
Step 4: Substitute the known values into the expression for Ksp and solve
for Ksp.
Ksp = (4.53 ×10−3)(9.06 ×10−3)2= 3.91 ×10−7
Therefore, the solubility product constant of lead(II) chloride is 3.91 ×10−7.
Question 14
Question
A student is conducting an experiment where they mix 50 mL of 0.1 M silver
nitrate solution with 50 mL of 0.1 M sodium chloride solution. Both solutions
are at room temperature. Calculate the mass of silver chloride precipitate that
will form. (Given: atomic masses of Ag = 107.87 g/mol, N = 14.01 g/mol, Na
= 22.99 g/mol, Cl = 35.45 g/mol)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between silver nitrate and sodium chloride:
AgNO3(aq) + NaCl(aq)→AgCl(s) + N aNO3(aq)
Step 2: Determine the limiting reactant by calculating the number of moles
of each reactant. Number of moles of silver nitrate:
nAgNO3=M×V= 0.1 M ×0.05 L = 0.005 mol
Number of moles of sodium chloride:
nNaCl =M×V= 0.1 M ×0.05 L = 0.005 mol
Step 3: Determine the limiting reactant by comparing the number of moles
of each reactant. Since both reactants have the same number of moles, the
limiting reactant is silver nitrate.
11
Step 4: Calculate the mass of silver chloride precipitate formed using the
stoichiometry of the reaction. From the balanced chemical equation, the mole
ratio of AgNO3 to AgCl is 1:1. The molar mass of AgCl:
MAgCl =Ag +Cl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Mass of AgCl formed:
mAgCl =nAgN O3×MAgCl = 0.005 mol ×143.32 g/mol = 0.717 g
Therefore, the mass of silver chloride precipitate that will form is 0.717 g.
Question 15
Question
Calculate the solubility of lead(II) chloride in water at 25
°
C. The solubility
product constant (Ksp) for lead(II) chloride is 1.7×10−5.
Solution
Step 1: Write the balanced equation for the dissociation of lead(II) chloride.
Step 2: Set up an ICE (Initial, Change, Equilibrium) table. Step 3: Define
the variables and the expression for the solubility of lead(II) chloride. Step 4:
Substitute the variables into the expression and solve for the solubility.
Step 1: The balanced equation for the dissociation of lead(II) chloride is:
PbCl2(s)⇌Pb2+(aq) + 2Cl−(aq)
Step 2: Set up the ICE table:
Substance PbCl2(s) Pb2+(aq) Cl−(aq)
Initial (M) s0 0
Change (M) −x+x+2x
Equilibrium (M) s−x x 2x
Step 3: Let xbe the solubility of lead(II) chloride. The expression for the
solubility product constant is:
Ksp = [Pb2+][Cl−]2=x(2x)2= 4x3
Step 4: Substitute the given Ksp value into the equation and solve for x:
1.7×10−5= 4x3
x3=1.7×10−5
4= 4.25 ×10−6
x=3
p4.25 ×10−6≈0.0176 M
Therefore, the solubility of lead(II) chloride in water at 25
°
C is approximately
0.0176 M.
12
Question 16
Question
Calculate the concentration of sulfate ions (SO2−
4) in a solution prepared by
mixing 150.0 mL of 0.200 M barium nitrate (Ba(N O3)2) with 300.0 mL of
0.100 M sodium sulfate (Na2SO4). Assume that the barium sulfate (BaSO4)
formed is insoluble and will precipitate out completely.
Solution
Step 1: Write the balanced chemical equation for the reaction between barium
nitrate and sodium sulfate:
Ba(N O3)2(aq) + N a2SO4(aq)→BaSO4(s)+2NaN O3(aq)
Step 2: Determine the limiting reagent in the reaction. - Calculate the
moles of barium nitrate (Ba(N O3)2) and sodium sulfate (Na2SO4): Moles of
Ba(N O3)2= 0.200 M ×0.150 L = 0.0300 mol Moles of N a2SO4= 0.100 M ×
0.300 L = 0.0300 mol
- Since the moles of both reactants are equal, either can be the limiting
reagent. Let’s assume that Ba(N O3)2is the limiting reagent.
Step 3: Calculate the moles of sulfate ions formed. - From the balanced
chemical equation, 1 mole of Ba(N O3)2reacts with 1 mole of SO2−
4. - Therefore,
moles of SO2−
4produced = moles of Ba(N O3)2= 0.0300 mol
Step 4: Calculate the total volume of the solution. Total volume = 150.0
mL + 300.0 mL = 450.0 mL = 0.450 L
Step 5: Calculate the concentration of sulfate ions. Concentration of sulfate
ions = moles of SO2−
4/ total volume Concentration of sulfate ions = 0.0300
mol / 0.450 L = 0.0667 M
Therefore, the concentration of sulfate ions in the solution is 0.0667 M.
Question 17
Question
Calculate the concentration of chloride ions (Cl−) in a solution if 35.6 grams of
silver chloride (AgCl) precipitate out when silver nitrate (AgNO3) is added to
the solution. Assume that all the Cl−ions in the solution react with the Ag+
ions to form AgCl.
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and chloride ions:
AgNO3+ Cl−→AgCl + NO−
3
13
Step 2: Calculate the molar mass of AgCl to determine the number of moles
of AgCl formed. Molar mass of AgCl = mass of Ag + mass of Cl Molar mass of
AgCl = 108 + 35.5 = 143.5 g/mol
Number of moles of AgCl = mass of AgCl
molar mass of AgCl =35.6 g
143.5 g/mol = 0.248 mol
Step 3: From the balanced chemical equation, we see that 1 mole of AgCl
is formed when 1 mole of Cl−reacts. Therefore, the number of moles of Cl−in
the solution is 0.248 mol.
Step 4: Calculate the concentration of Cl−ions in the solution. Concentra-
tion of Cl−ions = moles of Cl−
volume of solution
Since the volume of the solution is not given, the concentration cannot be
determined without additional information.
Question 18
Question
A student is conducting a precipitation experiment where they mix 50 ml of
a 0.2 M solution of barium chloride with 75 ml of a 0.1 M solution of sodium
sulfate. Calculate the maximum mass of barium sulfate that can be precipitated
from this solution.
(Density of barium sulfate = 4.5 g/cm3)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between barium chloride and sodium sulfate:
BaCl2+ Na2SO4→BaSO4+ 2NaCl
Step 2: Determine the limiting reactant by calculating the number of moles
of barium chloride and sodium sulfate:
Moles of BaCl2= Volume ×Concentration
= 0.050 L ×0.2 mol/L
= 0.010 mol
Moles of Na2SO4= Volume ×Concentration
= 0.075 L ×0.1 mol/L
= 0.0075 mol
Since barium chloride is the limiting reactant (0.010 mol ¡ 0.0075 mol), we
will use it to calculate the mass of barium sulfate precipitated.
14
Step 3: Calculate the theoretical yield of barium sulfate in grams:
Moles of BaSO4= 0.010 mol
Molar mass of BaSO4= 137.3 g/mol + 32.1 g/mol + (4 ×16.0 g/mol)
= 233.4 g/mol
Mass of BaSO4= Moles ×Molar mass
= 0.010 mol ×233.4 g/mol
= 2.334 g
Step 4: Calculate the maximum mass of barium sulfate that can be precipi-
tated:
Volume of BaSO4= Molarity ×Volume ×Density
= 0.2 mol/L ×0.050 L ×4.5 g/cm3
= 0.045 g
Therefore, the maximum mass of barium sulfate that can be precipitated
from this solution is 2.334 g.
Question 19
Question
A chemist is performing a precipitation reaction by mixing 50.0 mL of a 0.200
M calcium chloride (CaCl2) solution with 75.0 mL of a 0.150 M sodium sulfate
(Na2SO4) solution. Assuming complete precipitation, calculate the mass of
CaSO4that will form.
(Hint: Consider the stoichiometry of the reaction and the limiting reactant.)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between CaCl2and Na2SO4:
CaCl2+ Na2SO4→CaSO4+ 2NaCl
Step 2: Determine the limiting reactant by calculating the moles of each
reactant.
For CaCl2solution:
Moles of CaCl2= Volume (L) ×Molarity = 0.0500 L ×0.200 M = 0.0100 mol
For Na2SO4solution:
Moles of Na2SO4= Volume (L) ×Molarity = 0.0750 L ×0.150 M = 0.0113 mol
15
Since CaCl2has fewer moles than Na2SO4, CaCl2is the limiting reactant.
Step 3: Calculate the theoretical yield of CaSO4based on the limiting reac-
tant.
From the balanced chemical equation, 1 mol of CaCl2produces 1 mol of
CaSO4. Therefore, 0.0100 mol of CaCl2will produce 0.0100 mol of CaSO4.
Step 4: Calculate the mass of CaSO4formed using the molar mass of CaSO4.
Molar mass of CaSO4= 40.1 g/mol + 32.1 g/mol + 4(16.0 g/mol) = 136.1 g/mol
Mass of CaSO4= Moles ×Molar mass = 0.0100 mol ×136.1 g/mol = 1.36 g
Therefore, the mass of CaSO4that will form is 1.36 g.
Question 20
Question
A solution contains 0.2 M lead(II) nitrate and 0.1 M potassium iodide. Calculate
the concentration of lead(II) iodide that forms at the instant the reaction begins.
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and potassium iodide to form lead(II) iodide and potassium nitrate.
Pb(NO3)2(aq)+ 2KI(aq)→PbI2(s)+ 2KNO3(aq)
Step 2: Determine the limiting reactant by calculating the moles of each reac-
tant. Given:
[Pb(NO3)2]=0.2 M
[KI] = 0.1 M
Since the stoichiometric ratio between lead(II) nitrate and lead(II) iodide is 1:1,
and between potassium iodide and lead(II) iodide is 2:1, the limiting reactant
will be lead(II) nitrate.
Step 3: Calculate the concentration of lead(II) iodide that forms. Since
lead(II) nitrate is the limiting reactant, all of it will react to form lead(II)
iodide. Thus, the concentration of lead(II) iodide formed will be equal to the
initial concentration of lead(II) nitrate:
[PbI2] = 0.2 M
Therefore, at the instant the reaction begins, the concentration of lead(II)
iodide that forms is 0.2 M.
16
Question 21
Question
A solution was prepared by dissolving 10.0 g of calcium chloride (CaCl2) in 50.0
mL of water. If the solubility product constant of CaCl2is 3.9×10−6mol2/L2,
will a precipitate form when 25.0 mL of 0.030 M sodium chloride (NaCl) solution
is added to this solution? (Assume the volume change is negligible)
Solution
Step 1: Write the balanced chemical equation for the dissociation of CaCl2and
NaCl.
CaCl2(s)⇌Ca2+(aq) + 2Cl−(aq)
NaCl(s)⇌Na+(aq) + Cl−(aq)
Step 2: Calculate the initial concentrations of Ca2+ and Cl−ions in the
solution before adding NaCl.
The initial concentration of Ca2+ ions:
[Ca2+] = moles of Ca2+
total volume of solution in L =10.0 g
110.0 g/mol ×0.050 L = 18.2 mol/L
The initial concentration of Cl−ions:
[Cl−]=2×[Ca2+] = 36.4 mol/L
Step 3: Calculate the concentration of Cl−ions after adding NaCl.
The concentration of Cl−ions after adding NaCl:
[Cl−]final =[NaCl] ×volume of NaCl solution
total volume of solution =0.030 mol/L ×0.025 L
0.075 L = 0.010 mol/L
Step 4: Calculate the ion product, Q, for CaCl2.
Q= [Ca2+]×[Cl−]2= 18.2 mol/L ×(36.4 mol/L)2= 24,928
Step 5: Compare Qto the solubility product constant Ksp.
Since Q>Ksp, a precipitate of CaCl2will form when 25.0 mL of 0.030 M
sodium chloride solution is added to the original solution.
Question 22
Question
Calculate the concentration of a new precipitate that forms when 100.0 mL of
0.10 M copper(II) chloride reacts completely with excess sodium hydroxide to
form copper(II) hydroxide. The equation for the reaction is:
CuCl2(aq) + 2NaOH(aq)→Cu(OH)2(s)+2NaCl(aq)
17
Solution
Step 1: Determine the limiting reactant.
From the balanced chemical equation, we see that 1 mole of copper(II) chloride
reacts with 2 moles of sodium hydroxide.
Number of moles of CuCl2= 0.100 M ×0.100 L = 0.0100 mol
Number of moles of NaOH = 2 ×0.0100 = 0.0200 mol
Step 2: Calculate the amount of copper(II) hydroxide formed.
From the balanced chemical equation, 1 mole of copper(II) chloride forms 1
mole of copper(II) hydroxide.
Number of moles of Cu(OH)2formed = 0.0100 mol
Step 3: Calculate the concentration of copper(II) hydroxide.
Volume of solution = 100.0 mL = 0.100 L
Concentration of Cu(OH)2=0.0100 mol
0.100 L = 0.10 M
Question 23
Question
A chemist is performing a precipitation reaction by combining 50.0 mL of 0.100
M CaCl2 with 75.0 mL of 0.150 M Na3PO4 in a reaction vessel. The balanced
chemical equation for the reaction is:
3CaCl2(aq)+2Na3P O4(aq)→Ca3(P O4)2(s)+6NaCl(aq)
What is the maximum mass of calcium phosphate, Ca3(P O4)2, that can be
formed in grams?
Solution
Step 1: Calculate the moles of CaCl2 and Na3PO4 used in the reaction.
Moles of CaCl2 = Volume ×Molarity
Moles of CaCl2 = 50.0×10−3L×0.100 mol/L = 5.00 ×10−3mol
Moles of Na3PO4 = Volume ×Molarity
Moles of Na3PO4 = 75.0×10−3L×0.150 mol/L = 1.13 ×10−2mol
Step 2: Determine the limiting reactant by comparing the moles of each
reactant to their stoichiometric coefficients in the balanced equation.
Moles of Ca3(PO4)2 produced = 5.00 ×10−3mol
3= 1.67 ×10−3mol
Moles of Ca3(PO4)2 produced = 1.13 ×10−2mol
2= 5.65 ×10−3mol
18
Since CaCl2 produces fewer moles of Ca3(P O4)2and is therefore the limiting
reactant.
Step 3: Calculate the mass of Ca3(P O4)2that can be formed using the
limiting reactant.
Molar mass of Ca3(P O4)2= 3 ×Atomic mass of Ca + 2 ×Atomic mass of P + 8 ×Atomic mass of O
Molar mass of Ca3(P O4)2= 3 ×40.08 + 2 ×30.97 + 8 ×16.00 = 310.18 g/mol
Mass of Ca3(P O4)2= Moles ×Molar mass
Mass of Ca3(P O4)2= 1.67 ×10−3mol ×310.18 g/mol = 0.518 g
Therefore, the maximum mass of calcium phosphate, Ca3(P O4)2, that can
be formed is 0.518 grams.
Question 24
Question
A student is performing an experiment to determine the concentration of chlo-
ride ions in a water sample. The student adds an excess of silver nitrate solution
to a 100.0 mL sample of the unknown water sample. The precipitate formed is
filtered, dried, and found to have a mass of 0.385 g. Calculate the concentration
of chloride ions in the water sample in parts per million (ppm).
(Molar mass of AgCl = 143.32 g/mol)
Solution
Step 1: Determine the mass of chloride ions in the precipitate. The balanced
chemical equation for the reaction between silver nitrate (AgNO3) and chloride
ions (Cl−) is:
AgNO3+ Cl−→AgCl + NO−
3
Since the mass of the precipitate is 0.385 g and the molar mass of AgCl is
143.32 g/mol, we can calculate the moles of AgCl formed:
Moles of AgCl = Mass of AgCl
Molar mass of AgCl =0.385 g
143.32 g/mol = 0.00268 mol
Step 2: Calculate the moles of chloride ions in the water sample. From the
balanced chemical equation, we see that 1 mole of AgCl precipitate is formed for
every mole of chloride ions. Therefore, the moles of chloride ions in the water
sample is also 0.00268 mol.
Step 3: Calculate the concentration of chloride ions in the water sample in
ppm. The water sample has a volume of 100.0 mL, which is equivalent to 0.1000
L. Therefore, the concentration of chloride ions in the water sample is:
Concentration of Cl−=Moles of Cl−
Volume of water sample in L =0.00268 mol
0.1000 L = 0.0268 M
19
To convert this concentration to parts per million (ppm), we use the rela-
tionship:
ppm = Concentration ×106
Therefore, the concentration of chloride ions in the water sample is 26.8
ppm.
Question 25
Question
Calculate the mass of barium sulfate that can be formed when 250.0 mL of 0.150
M barium chloride solution is mixed with excess sulfuric acid. The balanced
chemical equation for the reaction is:
BaCl2(aq)+H2SO4(aq)→BaSO4(s) + 2HCl(aq)
Solution
Step 1: Write the balanced chemical equation for the reaction.
BaCl2(aq)+H2SO4(aq)→BaSO4(s) + 2HCl(aq)
Step 2: Determine the moles of barium chloride used. Given volume of
barium chloride solution = 250.0 mL = 0.250 L Molarity of barium chloride
solution = 0.150 M
Moles of BaCl2= Volume ×Molarity = 0.250 L ×0.150 mol/L = 0.0375 mol
Step 3: Use the stoichiometry of the balanced equation to find the moles of
barium sulfate formed. According to the balanced chemical equation, 1 mole of
barium chloride reacts with 1 mole of barium sulfate.
Moles of BaSO4= Moles of BaCl2= 0.0375 mol
Step 4: Calculate the mass of barium sulfate formed. Molar mass of BaSO
= 137.3 g/mol (Ba) + 32.1 g/mol (S) + (4 * 16.0 g/mol) = 233.4 g/mol
Mass of BaSO4= Moles ×Molar mass = 0.0375 mol ×233.4 g/mol = 8.76 g
Therefore, the mass of barium sulfate that can be formed is 8.76 g.
20
Question 26
Question
Calculate the mass of barium sulfate (BaSO4) that will precipitate when a
solution containing 5.00 g of barium chloride (BaCl2) is mixed with a solution
containing 10.0 g of sodium sulfate (Na2SO4). Assume that the reaction goes
to completion.
Solution
Step 1: Write the balanced chemical equation for the reaction between barium
chloride and sodium sulfate.
BaCl2+ Na2SO4→BaSO4+ 2NaCl
Step 2: Calculate the molar masses of BaCl2and Na2SO4. Molar mass of
BaCl2:
1×Ba + 2 ×Cl = 137.33 g/mol
Molar mass of Na2SO4:
2×Na + 1 ×S+4×O = 142.04 g/mol
Step 3: Determine the limiting reactant. Calculate the moles of each reac-
tant:
moles of BaCl2=5.00 g
137.33 g/mol = 0.0364 mol
moles of Na2SO4=10.0 g
142.04 g/mol = 0.0705 mol
Since 0.0364 moles of BaCl2can only react with 0.0182 moles of Na2SO4,
Na2SO4is the limiting reactant.
Step 4: Calculate the mass of BaSO4produced. The molar mass of BaSO4
is:
1×Ba + 1 ×S+4×O = 233.39 g/mol
Mass of BaSO4produced:
0.0705 mol ×233.39 g/mol = 16.47 g
Therefore, 16.47 g of barium sulfate will precipitate when 5.00 g of barium
chloride is mixed with 10.0 g of sodium sulfate.
Question 27
Question
A solution contains 0.3 M copper(II) nitrate, 0.2 M sodium sulfide, and 0.1 M
sulfuric acid. How many grams of copper(II) sulfide will precipitate when 50.0
mL of the solution is mixed with 50.0 mL of 0.1 M sodium sulfide? (Assume all
volumes are additive and the reaction goes to completion.)
21
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between copper(II) nitrate and sodium sulfide. The balanced chemical equation
is:
Cu(N O3)2 + Na2S→CuS + 2NaNO3
Step 2: Determine the limiting reactant. Calculate the moles of each reac-
tant: - Moles of Cu(NO3)2 in 50.0 mL of 0.3 M solution:
(0.3 M) ×(0.050 L) = 0.015 moles Cu(NO3)2
- Moles of Na2S in 50.0 mL of 0.1 M solution:
(0.1 M) ×(0.050 L) = 0.005 moles Na2S
Since 1 mole of Cu(NO3)2 reacts with 1 mole of Na2S, Na2S is the limiting
reactant.
Step 3: Calculate the moles of CuS that will precipitate. From the balanced
chemical equation, 1 mole of Cu(NO3)2 produces 1 mole of CuS:
0.005 moles CuS will precipitate
Step 4: Calculate the mass of CuS precipitated. - Calculate the molar mass
of CuS:
Cu : 63.5 g/mol and S: 32.1 g/mol
Molar mass = 63.5 + 32.1 = 95.6 g/mol
- Mass of CuS precipitated:
0.005 moles ×95.6 g/mol = 0.478 grams CuS
Therefore, 0.478 grams of copper(II) sulfide will precipitate when 50.0 mL
of the solution is mixed with 50.0 mL of 0.1 M sodium sulfide.
Question 28
Question
Calculate the solubility of silver chloride (AgCl) in water at 25◦C. Given that
the Ksp of AgCl is 1.8×10−10.
Solution
Step 1: Write the dissociation of AgCl:
AgCl(s)⇌Ag+(aq) + Cl−(aq)
22
Step 2: Write the expression for the solubility product constant (Ksp):
Ksp = [Ag+][Cl−]
Step 3: Let xbe the solubility of AgCl in moles per liter:
AgCl(s)⇌xmol/L + xmol/L
Step 4: Substitute the solubility values into the Ksp expression:
1.8×10−10 =x×x
x2= 1.8×10−10
x=p1.8×10−10
Step 5: Calculate the solubility of AgCl:
x=p1.8×10−10
x≈1.3×10−5mol/L
Therefore, the solubility of AgCl in water at 25◦C is approximately 1.3×10−5
mol/L.
Question 29
Question
A solution contains 0.1 M lead(II) nitrate (Pb(NO3)2) and 0.1 M sodium sulfate
(Na2SO4). Will precipitation occur if these two solutions are mixed? If so, what
is the chemical equation for the reaction, and what is the concentration of the
remaining ions in solution after the reaction reaches completion?
Solution
Step 1: Determine the possible precipitate by considering the ions in the two
solutions. The possible precipitate is lead(II) sulfate (PbSO4) because Pb2+
ions from lead(II) nitrate and SO2−
4ions from sodium sulfate can react to form
an insoluble lead sulfate precipitate.
Step 2: Write the balanced chemical equation for the precipitation reaction:
Pb(NO3)2+ Na2SO4→PbSO4+ 2NaNO3
Step 3: Determine the concentration of each ion in the solution after the
reaction reaches completion. Initially, the concentration of Pb2+ and SO2−
4ions
is 0.1 M each. After the reaction, all the Pb2+ ions will react with SO2−
4ions to
form PbSO4, which is insoluble and precipitates out of the solution. Therefore,
all Pb2+ ions will be consumed in the reaction, leaving no remaining Pb2+ ions
in solution. The remaining SO2−
4ion concentration will be the excess after all
the Pb2+ ions have reacted.
Therefore, the concentration of remaining SO2−
4ions in the solution is 0.1
M.
23
Question 30
Question
A solution contains 0.1 M of silver nitrate (AgNO3) and 0.2 M of potassium
chloride (KCl). What is the maximum concentration of silver chloride (AgCl)
that can be precipitated from this solution? (Ksp of AgCl is 1.8×10−10)
Solution
Step 1: Write the balanced chemical equation for the precipitation of silver
chloride:
AgNO3+ KCl →AgCl + KNO3
Step 2: Calculate the initial concentration of ions: Given: [Ag] = 0.1 M [Cl]
= 0.2 M
Step 3: Determine the limiting reactant to find the maximum concentration
of silver chloride that can be precipitated: From the balanced chemical equation,
we see that 1 mole of AgCl is formed from 1 mole of Ag and 1 mole of Cl.
Step 4: Calculate the concentration of silver ions and chloride ions formed
by the reaction: The maximum amount of AgCl precipitated will be limited by
the reactant that produces less AgCl. This can be determined by comparing
the initial concentrations of silver ions and chloride ions with the stoichiometry
of the reaction.
Step 5: Calculate the concentration of silver ions remaining: Since Cl is in
excess, all of the AgNO3will be consumed. Therefore, the concentration of
silver ions remaining is 0 M.
Step 6: Calculate the maximum concentration of AgCl that can be pre-
cipitated: Using the concentration of silver ions remaining, we can find the
maximum concentration of AgCl that can be precipitated. This is done by
multiplying the concentration of the limiting ion by the stoichiometry of the
reaction.
Step 7: Calculate the Ksp expression for AgCl: The Ksp expression for AgCl
is given by:
Ksp = [Ag][Cl]
Step 8: Substitute the known values into the Ksp expression and solve for
the maximum concentration of AgCl.
Question 31
Question
Calculate the solubility of silver chloride (AgCl) in water at 25◦C. Given that
the Ksp of AgCl is 1.6×10−10.
24
Solution
Step 1: Write the equation for the dissociation of silver chloride:
AgCl(s)⇌Ag+(aq) + Cl−(aq)
Step 2: Write the expression for the solubility product (Ksp) of AgCl:
Ksp = [Ag+][Cl−]
Step 3: Let the solubility of AgCl be represented as x. Then at equilibrium,
the concentrations of Ag+and Cl−ions will also be x. Thus, substitute these
values into the Ksp expression:
1.6×10−10 =x×x
1.6×10−10 =x2
Step 4: Solve for x:
x=p1.6×10−10 = 1.26 ×10−5
Therefore, the solubility of silver chloride in water at 25◦Cis 1.26 ×10−5
mol/L.
Question 32
Question
Calculate the mass of lead(II) iodide (PbI2) that can be formed from the reaction
of 50.0 mL of 0.150 M lead(II) nitrate (Pb(NO3)2) with excess potassium iodide
(KI). The balanced chemical equation for the reaction is:
Pb(NO3)2(aq) + 2 KI(aq) →PbI2(s) + 2 KNO3(aq)
Solution
Step 1: Write the balanced chemical equation for the reaction.
Pb(NO3)2(aq) + 2 KI(aq) →PbI2(s) + 2 KNO3(aq)
Step 2: Determine the moles of lead(II) nitrate (Pb(NO3)2). Given: Volume
of Pb(NO3)2solution = 50.0 mL = 0.0500 L Molarity of Pb(NO3)2solution =
0.150 M
Moles of Pb(NO3)2= Molarity×Volume = 0.150 mol/L×0.0500 L = 0.00750 mol
Step 3: Use the mole ratio from the balanced equation to determine the
moles of PbI2formed. From the balanced equation, 1 mole of Pb(NO3)2reacts
with 1 mole of PbI2.
Moles of PbI2formed = 0.00750 mol
25
Step 4: Calculate the mass of PbI2formed. The molar mass of PbI2is the
sum of the molar masses of lead (Pb) and iodine (I) in the compound.
Molar mass of P bI2= molar mass of Pb + 2 ×molar mass of I
= 207.2 g/mol + 2 ×126.9 g/mol
= 459.2 g/mol
Mass of PbI2= Moles of PbI2×Molar mass of PbI2= 0.00750 mol×459.2 g/mol
= 3.44 g
Therefore, the mass of lead(II) iodide (PbI2) that can be formed is 3.44
grams.
Question 33
Question
Calculate the mass of precipitate that forms when 50.0 mL of 0.100 M silver
nitrate reacts with excess sodium chloride forming silver chloride, which has a
solubility product constant Ksp = 1.77 ×10−10.
Solution
Step 1: Write the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between silver nitrate (AgNO3)
and sodium chloride (NaCl) to form silver chloride precipitate (AgCl) is:
AgNO3+ NaCl →AgCl + NaNO3
Step 2: Determine the limiting reactant.
Using the balanced chemical equation, we can see that the molar ratio of AgNO3
to AgCl is 1:1. Therefore, all 0.100 M of AgNO3will react with NaCl to form
AgCl.
Step 3: Calculate the moles of AgCl formed.
Since the molar ratio of AgNO3to AgCl is 1:1, the moles of AgCl formed will
be equal to the moles of AgNO3used in the reaction.
Moles of AgCl = Moles of AgNO3= Volume ×Molarity
Moles of AgCl = 0.0500 L ×0.100 mol/L = 0.00500 mol
Step 4: Calculate the mass of AgCl formed.
The molar mass of AgCl is:
Ag: 107.87 g/mol, Cl: 35.45 g/mol
26
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
The mass of AgCl formed is:
Mass of AgCl = Moles of AgCl ×Molar mass of AgCl
Mass of AgCl = 0.00500 mol ×143.32 g/mol = 0.716 g
Therefore, the mass of precipitate (AgCl) that forms is 0.716 grams.
Question 34
Question
Calculate the mass of precipitate that forms when 100.0 mL of a 0.100 M solution
of silver nitrate, AgNO3, is mixed with 150.0 mL of a 0.150 M solution of sodium
chloride, NaCl. Assume the reaction goes to completion, forming solid silver
chloride, AgCl.
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium chloride:
AgNO3+ NaCl →AgCl + NaNO3
Step 2: Determine the limiting reactant. First, calculate the moles of silver
nitrate (AgNO3) and sodium chloride (NaCl) using the provided concentrations
and volumes: For silver nitrate:
moles of AgNO3= M ×V=0.100 mol/L ×0.100 L = 0.0100 mol
For sodium chloride:
moles of NaCl = M ×V = 0.150 mol/L ×0.150 L = 0.0225 mol
Step 3: Determine the limiting reactant. The stoichiometry of the balanced
chemical equation shows that 1 mole of AgNO3reacts with 1 mole of NaCl to
form 1 mole of AgCl. Therefore, the limiting reactant is the one that produces
the least amount of AgCl.
In this case, AgNO3produces only 0.0100 moles of AgCl while NaCl can
produce 0.0225 moles of AgCl. Thus, AgNO3is the limiting reactant.
Step 4: Calculate the mass of precipitate formed. The molar mass of AgCl
is approximately 143.32 g/mol.
Mass of AgCl = moles of AgCl ×molar mass of AgCl
Mass of AgCl = 0.0100 mol ×143.32 g/mol = 1.4330 g
Therefore, the mass of precipitate that forms when 100.0 mL of a 0.100 M
solution of silver nitrate is mixed with 150.0 mL of a 0.150 M solution of sodium
chloride is approximately 1.4330 grams.
27
mass of AgCl = mol of AgCl ×molar mass of AgCl
The molar mass of AgCl is the sum of the atomic masses of silver (Ag) and
chlorine (Cl):
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Therefore,
mass of AgCl = 0.010 mol ×143.32 g/mol = 1.4332 g
The mass of silver chloride that can be formed when 50.0 mL of 0.200 M
silver nitrate solution is mixed with 75.0 mL of 0.150 M sodium chloride solution
is 1.4332 g.
Question 2
Question
A certain chemical reaction produces a precipitate when two solutions are mixed.
The balanced chemical equation is given by:
BaCl2(aq) + Na2SO4(aq)→BaSO4(s) + 2NaCl(aq)
If 500.0 mL of a 0.200 M solution of BaCl2 is mixed with 300.0 mL of a
0.150 M solution of Na2SO4, what mass of BaSO4 will be produced? (Assume
that all of the BaCl2 and Na2SO4 react to produce BaSO4.)
Solution
Step 1: Calculate the moles of BaCl2 and Na2SO4 in the solutions. Given:
Volume of BaCl2solution = 500.0 mL = 0.5000 L
Molarity of BaCl2solution = 0.200 M
Volume of Na2SO4solution = 300.0 mL = 0.3000 L
Molarity of Na2SO4solution = 0.150 M
Moles of BaCl2:
Moles of BaCl2= Molarity ×Volume in Liters
Moles of BaCl2= 0.200 mol/L ×0.5000 L
Moles of BaCl2= 0.1000 mol
2
Moles of Na2SO4:
Moles of Na2SO4= Molarity ×Volume in Liters
Moles of Na2SO4= 0.150 mol/L ×0.3000 L
Moles of Na2SO4= 0.0450 mol
Step 2: Determine the limiting reactant. From the balanced chemical equa-
tion, we can see that 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to produce
1 mole of BaSO4. Since the mole ratio is 1:1, the limiting reactant will be the
one that is completely consumed first in the reaction.
Initially, we have: - Moles of BaCl2: 0.1000 mol - Moles of Na2SO4: 0.0450
mol
To find the limiting reactant, we compare the moles of product that each
reactant could produce. Since the mole ratio between BaCl2 and Na2SO4 is
1:1: 1. For BaCl2: Moles of BaSO4 produced = 0.1000 mol 2. For Na2SO4:
Moles of BaSO4 produced = 0.0450 mol
Since Na2SO4 produces fewer moles of BaSO4, Na2SO4 is the limiting re-
actant.
Step 3: Calculate the mass of BaSO4 produced. Given:
Molar mass of BaSO4= 137.3 g/mol
Moles of BaSO4 produced will be the same as moles of Na2SO4 used in the
reaction:
Moles of BaSO4= 0.0450 mol
Finally, we calculate the mass of BaSO4 produced:
Mass of BaSO4= Moles of BaSO4×Molar mass of BaSO4
Mass of BaSO4= 0.0450 mol ×137.3 g/mol
Mass of BaSO4= 6.18 g
Therefore, the mass of BaSO4 produced when 500.0 mL of a 0.200 M solution
of BaCl2 is mixed with 300.
Question 3
Question
A solution is prepared by mixing 150 mL of a 0.2 M lead(II) nitrate solution with
200 mL of a 0.3 M potassium iodide solution. Calculate the mass of lead(II)
iodide that precipitates. (Assume the reaction goes to completion)
3
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and potassium iodide. The balanced chemical equation for the reaction
is:
Pb(NO3)2(aq)+ 2KI(aq)→PbI2(s)+ 2KNO3
Step 2: Determine the limiting reactant. First, calculate the number of
moles of each reactant: For lead(II) nitrate (Pb(NO3)2): Number of moles =
Concentration (M) ×Volume (L) Number of moles = 0.2 mol/L ×0.150 L =
0.03 moles For potassium iodide (KI): Number of moles = Concentration (M)
×Volume (L) Number of moles = 0.3 mol/L ×0.200 L = 0.06 moles
Since we need 2 moles of KI for every mole of Pb(NO3)2, the limiting reactant
is Pb(NO3)2.
Step 3: Calculate the mass of lead(II) iodide precipitated. From the balanced
chemical equation, we see that 1 mole of Pb(NO3)2 produces 1 mole of PbI2.
The molar mass of PbI2 is 461 g/mol. Number of moles of PbI2 formed =
Number of moles of Pb(NO3)2 Number of moles of PbI2 = 0.03 moles Mass of
PbI2 formed = Number of moles of PbI2 ×Molar mass of PbI2 Mass of PbI2
formed = 0.03 moles ×461 g/mol = 13.83 g
Therefore, the mass of lead(II) iodide that precipitates is 13.83 grams.
Question 4
Question
Calculate the mass of barium sulfate (BaSO4) that precipitates when 50.0 mL
of 0.200 M barium chloride (BaCl2) reacts with excess sodium sulfate (Na2SO4)
according to the following balanced chemical equation:
BaCl2+ Na2SO4−→ BaSO4+ 2NaCl
Solution
Step 1: Write the balanced chemical equation and identify the limiting reactant.
The balanced chemical equation is given as:
BaCl2+ Na2SO4−→ BaSO4+ 2NaCl
This equation indicates that 1 mole of BaCl2reacts with 1 mole of Na2SO4to
produce 1 mole of BaSO4.
Given that the volume of BaCl2is 50.0 mL and the concentration is 0.200
M, we need to calculate the number of moles of BaCl2:
moles of BaCl2= concentration×volume = 0.200 mol/L×0.0500 L = 0.0100 mol
Since the ratio of BaCl2to Na2SO4is 1:1, the number of moles of Na2SO4
required to react with all the BaCl2is also 0.0100 mol.
4
Step 2: Calculate the mass of BaSO4precipitated. From the balanced chem-
ical equation, we see that 1 mole of BaSO4is produced for every 1 mole of BaCl2
reacted.
The molar mass of BaSO4is calculated as:
Ba = 137.33 g/mol
S = 32.07 g/mol
4×O = 4 ×16.00 g/mol = 64.00 g/mol
Adding these up, we get:
137.33 + 32.07 + 64.00 = 233.40 g/mol
The mass of BaSO4that precipitates can be calculated as:
mass = moles ×molar mass = 0.0100 mol ×233.40 g/mol = 2.33 g
Therefore, the mass of barium sulfate (BaSO4) that precipitates is 2.33 g.
Question 5
Question
Calculate the molarity of MgCl2in a solution if 3.50 grams of MgCl2is dissolved
in enough water to make 250.0 mL of solution.
(Given: molar mass of MgCl2= 95.21 g/mol)
Solution
Step 1: Calculate the number of moles of MgCl2.
moles of MgCl2=mass
molar mass =3.50 g
95.21 g/mol
Step 2: Convert the solution volume to liters.
Volume of solution = 250.0 mL ×1 L
1000 mL
Step 3: Calculate the molarity of MgCl2.
Molarity = moles
volume in liters =3.50 g/95.21 g/mol
250.0 mL/1000 mL/L
5
Question 6
Question
A solution is prepared by dissolving 25.0 g of silver nitrate (AgNO3) in enough
water to make 0.250 L of solution. To this, excess sodium chloride (NaCl) is
added resulting in the precipitation of silver chloride (AgCl). If the mass of the
precipitate formed is 15.0 g, what is the concentration of silver ions (Ag+) in
the solution expressed in mol/L? (Assume complete precipitation of silver as
silver chloride)
(Hint: The balanced chemical equation for the precipitation of AgCl is
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq))
Solution
Step 1: Calculate the moles of AgNO3in the solution. Given mass of AgNO3:
25.0 g Molar mass of AgNO3: 107.87 g/mol + 14.01 g/mol + 3(16.00 g/mol) =
169.87 g/mol
Number of moles of AgNO3=25.0 g
169.87 g/mol = 0.1471 mol
Step 2: Determine the moles of Ag in the solution. From the balanced
equation, 1 mole of AgNO3produces 1 mole of Ag ions. Therefore, moles of
Ag+= 0.1471 mol
Step 3: Calculate the concentration of Ag+ions. Volume of solution = 0.250
L Concentration of Ag+=moles of Ag+
volume of solution =0.1471 mol
0.250L= 0.5884 mol/L
Therefore, the concentration of silver ions (Ag+) in the solution is 0.5884
mol/L.
Question 7
Question
A solution contains 0.1 M barium chloride and 0.2 M sodium sulfate. Calculate
the concentration of barium ion in the solution after precipitation of barium
sulfate.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between barium chloride and sodium sulfate.
BaCl2+ Na2SO4→BaSO4(s) + 2NaCl
Step 2: Determine the limiting reactant by comparing the number of moles
of BaCl2and Na2SO4.
Step 3: Calculate the number of moles of BaSO4formed using the stoichiom-
etry of the balanced equation.
6
Step 4: Since barium sulfate is insoluble, it will precipitate. Calculate the
concentration of Ba2+ ions in the solution after precipitation.
Step 5: Using the initial volume of the solution, calculate the final concen-
tration of Ba2+ ions.
Step 6: Express the concentration of Ba2+ ions in the solution.
Question 8
Question
A solution contains 25.0 g of potassium iodide (KI) in 150.0 mL of water. If
silver nitrate (AgNO3) is added to the solution, how many grams of silver iodide
(AgI) will precipitate out? Assume that the reaction goes to completion and
that only the formation of AgI is considered.
Solution
Step 1: Write the balanced chemical equation for the reaction between KI and
AgNO3to form AgI:
KI + AgNO3→AgI + KNO3
Step 2: Calculate the number of moles of KI in the solution:
Number of moles of KI = Mass (g)
Molar mass (g/mol) =25.0 g
166.00 g/mol = 0.1506 mol
Step 3: Find the limiting reagent by considering the stoichiometry of the
reaction. Since the reaction proceeds to completion, the limiting reagent is the
one that produces the least amount of AgI. Calculate the number of moles of
AgNO3needed to react with all of the KI:
Number of moles of AgNO3=Number of moles of KI
1= 0.1506 mol
Step 4: Calculate the mass of AgI produced:
Mass of AgI = Number of moles of AgI×Molar mass of AgI = 0.1506 mol×234.77 g/mol = 35.39 g
Therefore, 35.39 grams of silver iodide (AgI) will precipitate out.
Question 9
Question
A chemist is conducting a precipitation reaction in which 50.0 mL of 0.200 M
silver nitrate (AgNO3) is mixed with 75.0 mL of 0.150 M sodium chloride (NaCl).
Calculate the maximum mass of silver chloride (AgCl) that could precipitate.
(Given: molar mass of AgCl = 143.32 g/mol)
7
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between silver nitrate and sodium chloride:
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reactant by calculating the moles of each
reactant:
Moles of AgNO3:
Moles = Volume ×Molarity = 0.0500 L ×0.200 mol/L = 0.0100 mol
Moles of NaCl:
Moles = Volume ×Molarity = 0.0750 L ×0.150 mol/L = 0.0113 mol
Step 3: Use stoichiometry to determine the moles of AgCl that can be
formed: Since the reaction is 1:1 between AgNO3and AgCl, the moles of AgCl
formed will be 0.0100 mol (from AgNO3).
Step 4: Calculate the mass of AgCl formed using its molar mass:
Mass = Moles ×Molar mass = 0.0100 mol ×143.32 g/mol = 1.43 g
Therefore, the maximum mass of AgCl that could precipitate is 1.43 g.
Question 10
Question
A solution was prepared by mixing 50.0 mL of 0.200 M calcium chloride (CaCl2)
with 75.0 mL of 0.100 M sodium carbonate (Na2CO3). Calculate the mass of
the precipitate formed, assuming the reaction goes to completion.
Solution
Step 1: Write the balanced chemical equation for the reaction between calcium
chloride and sodium carbonate:
CaCl2+ Na2CO3→CaCO3+ 2NaCl
Step 2: Determine the limiting reagent by calculating the moles of each
reactant. For calcium chloride (CaCl2):
moles = molarity ×volume (L) = 0.200 mol/L ×0.0500 L = 0.0100 mol
For sodium carbonate (Na2CO3):
moles = molarity ×volume (L) = 0.100 mol/L ×0.0750 L = 0.00750 mol
8
Since 1 mole of CaCl2reacts with 1 mole of Na2CO3, Na2CO3is the limiting
reagent with 0.00750 mol.
Step 3: Calculate the theoretical yield of calcium carbonate (CaCO3) using
the limiting reagent:
moles of CaCO3= 0.00750 mol
Step 4: Convert the moles of calcium carbonate to grams using its molar
mass. The molar mass of CaCO3is:
1×Ca + 1 ×C+3×O = 40.08 + 12.01 + 3 ×16.00 = 100.09 g/mol
Therefore, the mass of CaCO3formed is:
mass = moles ×molar mass = 0.00750 mol ×100.09 g/mol = 0.751 g
Thus, the mass of the precipitate formed, calcium carbonate, is 0.751 g.
Question 11
Question
Calculate the mass of barium sulfate (BaSO4) that can be formed when 250.0
mL of a 0.200 M solution of barium chloride (BaCl2) is mixed with 300.0 mL
of a 0.150 M solution of sodium sulfate (Na2SO4). Assume the reaction goes to
completion.
Solution
Step 1: Write the balanced chemical equation for the reaction between barium
chloride and sodium sulfate to form barium sulfate and sodium chloride:
BaCl2+ Na2SO4→BaSO4+ 2NaCl
Step 2: Determine the limiting reactant by calculating the moles of each
reactant. The moles of barium chloride (BaCl2) is calculated as:
moles of BaCl2= Molarity ×Volume (L)
moles of BaCl2= 0.200 mol/L ×0.250 L = 0.0500 mol
The moles of sodium sulfate (Na2SO4) is calculated as:
moles of Na2SO4= Molarity ×Volume (L)
moles of Na2SO4= 0.150 mol/L ×0.300 L = 0.0450 mol
Since the stoichiometry of the reaction is 1:1 between BaCl2and Na2SO4,
Na2SO4is the limiting reactant.
9
Step 3: Calculate the theoretical yield of barium sulfate using the mole ratio
between Na2SO4and BaSO4.
moles of BaSO4= 0.0450 mol ×1 mol BaSO4
1 mol Na2SO4
= 0.0450 mol
Step 4: Calculate the mass of barium sulfate formed using its molar mass
(233.39 g/mol).
mass of BaSO4= 0.0450 mol ×233.39 g/mol = 10.50 g
Therefore, the mass of barium sulfate that can be formed is 10.50 g.
Question 12
Question
Calculate the concentration of chloride ions in a solution prepared by mixing
50.0 mL of a 0.200 M calcium chloride (CaCl2) solution with 100.0 mL of a
0.100 M sodium chloride (NaCl) solution. Assume the volumes are additive and
no volume changes occur upon mixing.
Solution
Step 1: Calculate the moles of chloride ions from each solute solution using the
formula n=C×V, where nis the number of moles, Cis the concentration,
and Vis the volume in liters.
For CaCl2:nCaCl2= 0.200 M ×0.0500 L = 0.010 mol
For NaCl: nNaCl = 0.100 M ×0.1000 L = 0.010 mol
Step 2: Determine the total moles of chloride ions in the solution by adding
the moles from both solutes. Total moles of chloride ions = nCaCl2+nNaCl =
0.010 mol + 0.010 mol = 0.020 mol
Step 3: Calculate the total volume of the solution. Vtotal = 0.0500 L +
0.1000 L = 0.1500 L
Step 4: Calculate the final concentration of chloride ions using the formula
Cfinal =ntotal
Vtotal .Cfinal =0.020 mol
0.1500 L = 0.133 M
Therefore, the concentration of chloride ions in the final solution is 0.133 M.
Question 13
Question
Calculate the solubility product constant (Ksp) of lead(II) chloride from the
solubility of lead(II) chloride in water at 25◦C, which is 4.53 x 10−3M.
10
Solution
Step 1: Write the balanced equation for the dissolution of lead(II) chloride in
water.
PbCl2(s)⇌Pb2+(aq) + 2Cl−(aq)
Step 2: Write the expression for the solubility product constant (Ksp).
Ksp = [Pb2+][Cl−]2
Step 3: List the known values from the question.
[Pb2+]=4.53 ×10−3M
[Cl−]=2×4.53 ×10−3= 9.06 ×10−3M
Step 4: Substitute the known values into the expression for Ksp and solve
for Ksp.
Ksp = (4.53 ×10−3)(9.06 ×10−3)2= 3.91 ×10−7
Therefore, the solubility product constant of lead(II) chloride is 3.91 ×10−7.
Question 14
Question
A student is conducting an experiment where they mix 50 mL of 0.1 M silver
nitrate solution with 50 mL of 0.1 M sodium chloride solution. Both solutions
are at room temperature. Calculate the mass of silver chloride precipitate that
will form. (Given: atomic masses of Ag = 107.87 g/mol, N = 14.01 g/mol, Na
= 22.99 g/mol, Cl = 35.45 g/mol)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between silver nitrate and sodium chloride:
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reactant by calculating the number of moles
of each reactant. Number of moles of silver nitrate:
nAgNO3=M×V= 0.1 M ×0.05 L = 0.005 mol
Number of moles of sodium chloride:
nNaCl =M×V= 0.1 M ×0.05 L = 0.005 mol
Step 3: Determine the limiting reactant by comparing the number of moles
of each reactant. Since both reactants have the same number of moles, the
limiting reactant is silver nitrate.
11
Step 4: Calculate the mass of silver chloride precipitate formed using the
stoichiometry of the reaction. From the balanced chemical equation, the mole
ratio of AgNO3 to AgCl is 1:1. The molar mass of AgCl:
MAgCl =Ag +Cl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Mass of AgCl formed:
mAgCl =nAgN O3×MAgCl = 0.005 mol ×143.32 g/mol = 0.717 g
Therefore, the mass of silver chloride precipitate that will form is 0.717 g.
Question 15
Question
Calculate the solubility of lead(II) chloride in water at 25
°
C. The solubility
product constant (Ksp) for lead(II) chloride is 1.7×10−5.
Solution
Step 1: Write the balanced equation for the dissociation of lead(II) chloride.
Step 2: Set up an ICE (Initial, Change, Equilibrium) table. Step 3: Define
the variables and the expression for the solubility of lead(II) chloride. Step 4:
Substitute the variables into the expression and solve for the solubility.
Step 1: The balanced equation for the dissociation of lead(II) chloride is:
PbCl2(s)⇌Pb2+(aq) + 2Cl−(aq)
Step 2: Set up the ICE table:
Substance PbCl2(s) Pb2+(aq) Cl−(aq)
Initial (M) s0 0
Change (M) −x+x+2x
Equilibrium (M) s−x x 2x
Step 3: Let xbe the solubility of lead(II) chloride. The expression for the
solubility product constant is:
Ksp = [Pb2+][Cl−]2=x(2x)2= 4x3
Step 4: Substitute the given Ksp value into the equation and solve for x:
1.7×10−5= 4x3
x3=1.7×10−5
4= 4.25 ×10−6
x=3
p4.25 ×10−6≈0.0176 M
Therefore, the solubility of lead(II) chloride in water at 25
°
C is approximately
0.0176 M.
12
Question 16
Question
Calculate the concentration of sulfate ions (SO2−
4) in a solution prepared by
mixing 150.0 mL of 0.200 M barium nitrate (Ba(N O3)2) with 300.0 mL of
0.100 M sodium sulfate (Na2SO4). Assume that the barium sulfate (BaSO4)
formed is insoluble and will precipitate out completely.
Solution
Step 1: Write the balanced chemical equation for the reaction between barium
nitrate and sodium sulfate:
Ba(N O3)2(aq) + N a2SO4(aq)→BaSO4(s)+2NaN O3(aq)
Step 2: Determine the limiting reagent in the reaction. - Calculate the
moles of barium nitrate (Ba(N O3)2) and sodium sulfate (Na2SO4): Moles of
Ba(N O3)2= 0.200 M ×0.150 L = 0.0300 mol Moles of N a2SO4= 0.100 M ×
0.300 L = 0.0300 mol
- Since the moles of both reactants are equal, either can be the limiting
reagent. Let’s assume that Ba(N O3)2is the limiting reagent.
Step 3: Calculate the moles of sulfate ions formed. - From the balanced
chemical equation, 1 mole of Ba(N O3)2reacts with 1 mole of SO2−
4. - Therefore,
moles of SO2−
4produced = moles of Ba(N O3)2= 0.0300 mol
Step 4: Calculate the total volume of the solution. Total volume = 150.0
mL + 300.0 mL = 450.0 mL = 0.450 L
Step 5: Calculate the concentration of sulfate ions. Concentration of sulfate
ions = moles of SO2−
4/ total volume Concentration of sulfate ions = 0.0300
mol / 0.450 L = 0.0667 M
Therefore, the concentration of sulfate ions in the solution is 0.0667 M.
Question 17
Question
Calculate the concentration of chloride ions (Cl−) in a solution if 35.6 grams of
silver chloride (AgCl) precipitate out when silver nitrate (AgNO3) is added to
the solution. Assume that all the Cl−ions in the solution react with the Ag+
ions to form AgCl.
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and chloride ions:
AgNO3+ Cl−→AgCl + NO−
3
13
Step 2: Calculate the molar mass of AgCl to determine the number of moles
of AgCl formed. Molar mass of AgCl = mass of Ag + mass of Cl Molar mass of
AgCl = 108 + 35.5 = 143.5 g/mol
Number of moles of AgCl = mass of AgCl
molar mass of AgCl =35.6 g
143.5 g/mol = 0.248 mol
Step 3: From the balanced chemical equation, we see that 1 mole of AgCl
is formed when 1 mole of Cl−reacts. Therefore, the number of moles of Cl−in
the solution is 0.248 mol.
Step 4: Calculate the concentration of Cl−ions in the solution. Concentra-
tion of Cl−ions = moles of Cl−
volume of solution
Since the volume of the solution is not given, the concentration cannot be
determined without additional information.
Question 18
Question
A student is conducting a precipitation experiment where they mix 50 ml of
a 0.2 M solution of barium chloride with 75 ml of a 0.1 M solution of sodium
sulfate. Calculate the maximum mass of barium sulfate that can be precipitated
from this solution.
(Density of barium sulfate = 4.5 g/cm3)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between barium chloride and sodium sulfate:
BaCl2+ Na2SO4→BaSO4+ 2NaCl
Step 2: Determine the limiting reactant by calculating the number of moles
of barium chloride and sodium sulfate:
Moles of BaCl2= Volume ×Concentration
= 0.050 L ×0.2 mol/L
= 0.010 mol
Moles of Na2SO4= Volume ×Concentration
= 0.075 L ×0.1 mol/L
= 0.0075 mol
Since barium chloride is the limiting reactant (0.010 mol ¡ 0.0075 mol), we
will use it to calculate the mass of barium sulfate precipitated.
14
Step 3: Calculate the theoretical yield of barium sulfate in grams:
Moles of BaSO4= 0.010 mol
Molar mass of BaSO4= 137.3 g/mol + 32.1 g/mol + (4 ×16.0 g/mol)
= 233.4 g/mol
Mass of BaSO4= Moles ×Molar mass
= 0.010 mol ×233.4 g/mol
= 2.334 g
Step 4: Calculate the maximum mass of barium sulfate that can be precipi-
tated:
Volume of BaSO4= Molarity ×Volume ×Density
= 0.2 mol/L ×0.050 L ×4.5 g/cm3
= 0.045 g
Therefore, the maximum mass of barium sulfate that can be precipitated
from this solution is 2.334 g.
Question 19
Question
A chemist is performing a precipitation reaction by mixing 50.0 mL of a 0.200
M calcium chloride (CaCl2) solution with 75.0 mL of a 0.150 M sodium sulfate
(Na2SO4) solution. Assuming complete precipitation, calculate the mass of
CaSO4that will form.
(Hint: Consider the stoichiometry of the reaction and the limiting reactant.)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between CaCl2and Na2SO4:
CaCl2+ Na2SO4→CaSO4+ 2NaCl
Step 2: Determine the limiting reactant by calculating the moles of each
reactant.
For CaCl2solution:
Moles of CaCl2= Volume (L) ×Molarity = 0.0500 L ×0.200 M = 0.0100 mol
For Na2SO4solution:
Moles of Na2SO4= Volume (L) ×Molarity = 0.0750 L ×0.150 M = 0.0113 mol
15
Since CaCl2has fewer moles than Na2SO4, CaCl2is the limiting reactant.
Step 3: Calculate the theoretical yield of CaSO4based on the limiting reac-
tant.
From the balanced chemical equation, 1 mol of CaCl2produces 1 mol of
CaSO4. Therefore, 0.0100 mol of CaCl2will produce 0.0100 mol of CaSO4.
Step 4: Calculate the mass of CaSO4formed using the molar mass of CaSO4.
Molar mass of CaSO4= 40.1 g/mol + 32.1 g/mol + 4(16.0 g/mol) = 136.1 g/mol
Mass of CaSO4= Moles ×Molar mass = 0.0100 mol ×136.1 g/mol = 1.36 g
Therefore, the mass of CaSO4that will form is 1.36 g.
Question 20
Question
A solution contains 0.2 M lead(II) nitrate and 0.1 M potassium iodide. Calculate
the concentration of lead(II) iodide that forms at the instant the reaction begins.
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and potassium iodide to form lead(II) iodide and potassium nitrate.
Pb(NO3)2(aq)+ 2KI(aq)→PbI2(s)+ 2KNO3(aq)
Step 2: Determine the limiting reactant by calculating the moles of each reac-
tant. Given:
[Pb(NO3)2]=0.2 M
[KI] = 0.1 M
Since the stoichiometric ratio between lead(II) nitrate and lead(II) iodide is 1:1,
and between potassium iodide and lead(II) iodide is 2:1, the limiting reactant
will be lead(II) nitrate.
Step 3: Calculate the concentration of lead(II) iodide that forms. Since
lead(II) nitrate is the limiting reactant, all of it will react to form lead(II)
iodide. Thus, the concentration of lead(II) iodide formed will be equal to the
initial concentration of lead(II) nitrate:
[PbI2] = 0.2 M
Therefore, at the instant the reaction begins, the concentration of lead(II)
iodide that forms is 0.2 M.
16
Question 21
Question
A solution was prepared by dissolving 10.0 g of calcium chloride (CaCl2) in 50.0
mL of water. If the solubility product constant of CaCl2is 3.9×10−6mol2/L2,
will a precipitate form when 25.0 mL of 0.030 M sodium chloride (NaCl) solution
is added to this solution? (Assume the volume change is negligible)
Solution
Step 1: Write the balanced chemical equation for the dissociation of CaCl2and
NaCl.
CaCl2(s)⇌Ca2+(aq) + 2Cl−(aq)
NaCl(s)⇌Na+(aq) + Cl−(aq)
Step 2: Calculate the initial concentrations of Ca2+ and Cl−ions in the
solution before adding NaCl.
The initial concentration of Ca2+ ions:
[Ca2+] = moles of Ca2+
total volume of solution in L =10.0 g
110.0 g/mol ×0.050 L = 18.2 mol/L
The initial concentration of Cl−ions:
[Cl−]=2×[Ca2+] = 36.4 mol/L
Step 3: Calculate the concentration of Cl−ions after adding NaCl.
The concentration of Cl−ions after adding NaCl:
[Cl−]final =[NaCl] ×volume of NaCl solution
total volume of solution =0.030 mol/L ×0.025 L
0.075 L = 0.010 mol/L
Step 4: Calculate the ion product, Q, for CaCl2.
Q= [Ca2+]×[Cl−]2= 18.2 mol/L ×(36.4 mol/L)2= 24,928
Step 5: Compare Qto the solubility product constant Ksp.
Since Q>Ksp, a precipitate of CaCl2will form when 25.0 mL of 0.030 M
sodium chloride solution is added to the original solution.
Question 22
Question
Calculate the concentration of a new precipitate that forms when 100.0 mL of
0.10 M copper(II) chloride reacts completely with excess sodium hydroxide to
form copper(II) hydroxide. The equation for the reaction is:
CuCl2(aq) + 2NaOH(aq)→Cu(OH)2(s)+2NaCl(aq)
17
Solution
Step 1: Determine the limiting reactant.
From the balanced chemical equation, we see that 1 mole of copper(II) chloride
reacts with 2 moles of sodium hydroxide.
Number of moles of CuCl2= 0.100 M ×0.100 L = 0.0100 mol
Number of moles of NaOH = 2 ×0.0100 = 0.0200 mol
Step 2: Calculate the amount of copper(II) hydroxide formed.
From the balanced chemical equation, 1 mole of copper(II) chloride forms 1
mole of copper(II) hydroxide.
Number of moles of Cu(OH)2formed = 0.0100 mol
Step 3: Calculate the concentration of copper(II) hydroxide.
Volume of solution = 100.0 mL = 0.100 L
Concentration of Cu(OH)2=0.0100 mol
0.100 L = 0.10 M
Question 23
Question
A chemist is performing a precipitation reaction by combining 50.0 mL of 0.100
M CaCl2 with 75.0 mL of 0.150 M Na3PO4 in a reaction vessel. The balanced
chemical equation for the reaction is:
3CaCl2(aq)+2Na3P O4(aq)→Ca3(P O4)2(s)+6NaCl(aq)
What is the maximum mass of calcium phosphate, Ca3(P O4)2, that can be
formed in grams?
Solution
Step 1: Calculate the moles of CaCl2 and Na3PO4 used in the reaction.
Moles of CaCl2 = Volume ×Molarity
Moles of CaCl2 = 50.0×10−3L×0.100 mol/L = 5.00 ×10−3mol
Moles of Na3PO4 = Volume ×Molarity
Moles of Na3PO4 = 75.0×10−3L×0.150 mol/L = 1.13 ×10−2mol
Step 2: Determine the limiting reactant by comparing the moles of each
reactant to their stoichiometric coefficients in the balanced equation.
Moles of Ca3(PO4)2 produced = 5.00 ×10−3mol
3= 1.67 ×10−3mol
Moles of Ca3(PO4)2 produced = 1.13 ×10−2mol
2= 5.65 ×10−3mol
18
Since CaCl2 produces fewer moles of Ca3(P O4)2and is therefore the limiting
reactant.
Step 3: Calculate the mass of Ca3(P O4)2that can be formed using the
limiting reactant.
Molar mass of Ca3(P O4)2= 3 ×Atomic mass of Ca + 2 ×Atomic mass of P + 8 ×Atomic mass of O
Molar mass of Ca3(P O4)2= 3 ×40.08 + 2 ×30.97 + 8 ×16.00 = 310.18 g/mol
Mass of Ca3(P O4)2= Moles ×Molar mass
Mass of Ca3(P O4)2= 1.67 ×10−3mol ×310.18 g/mol = 0.518 g
Therefore, the maximum mass of calcium phosphate, Ca3(P O4)2, that can
be formed is 0.518 grams.
Question 24
Question
A student is performing an experiment to determine the concentration of chlo-
ride ions in a water sample. The student adds an excess of silver nitrate solution
to a 100.0 mL sample of the unknown water sample. The precipitate formed is
filtered, dried, and found to have a mass of 0.385 g. Calculate the concentration
of chloride ions in the water sample in parts per million (ppm).
(Molar mass of AgCl = 143.32 g/mol)
Solution
Step 1: Determine the mass of chloride ions in the precipitate. The balanced
chemical equation for the reaction between silver nitrate (AgNO3) and chloride
ions (Cl−) is:
AgNO3+ Cl−→AgCl + NO−
3
Since the mass of the precipitate is 0.385 g and the molar mass of AgCl is
143.32 g/mol, we can calculate the moles of AgCl formed:
Moles of AgCl = Mass of AgCl
Molar mass of AgCl =0.385 g
143.32 g/mol = 0.00268 mol
Step 2: Calculate the moles of chloride ions in the water sample. From the
balanced chemical equation, we see that 1 mole of AgCl precipitate is formed for
every mole of chloride ions. Therefore, the moles of chloride ions in the water
sample is also 0.00268 mol.
Step 3: Calculate the concentration of chloride ions in the water sample in
ppm. The water sample has a volume of 100.0 mL, which is equivalent to 0.1000
L. Therefore, the concentration of chloride ions in the water sample is:
Concentration of Cl−=Moles of Cl−
Volume of water sample in L =0.00268 mol
0.1000 L = 0.0268 M
19
To convert this concentration to parts per million (ppm), we use the rela-
tionship:
ppm = Concentration ×106
Therefore, the concentration of chloride ions in the water sample is 26.8
ppm.
Question 25
Question
Calculate the mass of barium sulfate that can be formed when 250.0 mL of 0.150
M barium chloride solution is mixed with excess sulfuric acid. The balanced
chemical equation for the reaction is:
BaCl2(aq)+H2SO4(aq)→BaSO4(s) + 2HCl(aq)
Solution
Step 1: Write the balanced chemical equation for the reaction.
BaCl2(aq)+H2SO4(aq)→BaSO4(s) + 2HCl(aq)
Step 2: Determine the moles of barium chloride used. Given volume of
barium chloride solution = 250.0 mL = 0.250 L Molarity of barium chloride
solution = 0.150 M
Moles of BaCl2= Volume ×Molarity = 0.250 L ×0.150 mol/L = 0.0375 mol
Step 3: Use the stoichiometry of the balanced equation to find the moles of
barium sulfate formed. According to the balanced chemical equation, 1 mole of
barium chloride reacts with 1 mole of barium sulfate.
Moles of BaSO4= Moles of BaCl2= 0.0375 mol
Step 4: Calculate the mass of barium sulfate formed. Molar mass of BaSO
= 137.3 g/mol (Ba) + 32.1 g/mol (S) + (4 * 16.0 g/mol) = 233.4 g/mol
Mass of BaSO4= Moles ×Molar mass = 0.0375 mol ×233.4 g/mol = 8.76 g
Therefore, the mass of barium sulfate that can be formed is 8.76 g.
20
Question 26
Question
Calculate the mass of barium sulfate (BaSO4) that will precipitate when a
solution containing 5.00 g of barium chloride (BaCl2) is mixed with a solution
containing 10.0 g of sodium sulfate (Na2SO4). Assume that the reaction goes
to completion.
Solution
Step 1: Write the balanced chemical equation for the reaction between barium
chloride and sodium sulfate.
BaCl2+ Na2SO4→BaSO4+ 2NaCl
Step 2: Calculate the molar masses of BaCl2and Na2SO4. Molar mass of
BaCl2:
1×Ba + 2 ×Cl = 137.33 g/mol
Molar mass of Na2SO4:
2×Na + 1 ×S+4×O = 142.04 g/mol
Step 3: Determine the limiting reactant. Calculate the moles of each reac-
tant:
moles of BaCl2=5.00 g
137.33 g/mol = 0.0364 mol
moles of Na2SO4=10.0 g
142.04 g/mol = 0.0705 mol
Since 0.0364 moles of BaCl2can only react with 0.0182 moles of Na2SO4,
Na2SO4is the limiting reactant.
Step 4: Calculate the mass of BaSO4produced. The molar mass of BaSO4
is:
1×Ba + 1 ×S+4×O = 233.39 g/mol
Mass of BaSO4produced:
0.0705 mol ×233.39 g/mol = 16.47 g
Therefore, 16.47 g of barium sulfate will precipitate when 5.00 g of barium
chloride is mixed with 10.0 g of sodium sulfate.
Question 27
Question
A solution contains 0.3 M copper(II) nitrate, 0.2 M sodium sulfide, and 0.1 M
sulfuric acid. How many grams of copper(II) sulfide will precipitate when 50.0
mL of the solution is mixed with 50.0 mL of 0.1 M sodium sulfide? (Assume all
volumes are additive and the reaction goes to completion.)
21
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between copper(II) nitrate and sodium sulfide. The balanced chemical equation
is:
Cu(N O3)2 + Na2S→CuS + 2NaNO3
Step 2: Determine the limiting reactant. Calculate the moles of each reac-
tant: - Moles of Cu(NO3)2 in 50.0 mL of 0.3 M solution:
(0.3 M) ×(0.050 L) = 0.015 moles Cu(NO3)2
- Moles of Na2S in 50.0 mL of 0.1 M solution:
(0.1 M) ×(0.050 L) = 0.005 moles Na2S
Since 1 mole of Cu(NO3)2 reacts with 1 mole of Na2S, Na2S is the limiting
reactant.
Step 3: Calculate the moles of CuS that will precipitate. From the balanced
chemical equation, 1 mole of Cu(NO3)2 produces 1 mole of CuS:
0.005 moles CuS will precipitate
Step 4: Calculate the mass of CuS precipitated. - Calculate the molar mass
of CuS:
Cu : 63.5 g/mol and S: 32.1 g/mol
Molar mass = 63.5 + 32.1 = 95.6 g/mol
- Mass of CuS precipitated:
0.005 moles ×95.6 g/mol = 0.478 grams CuS
Therefore, 0.478 grams of copper(II) sulfide will precipitate when 50.0 mL
of the solution is mixed with 50.0 mL of 0.1 M sodium sulfide.
Question 28
Question
Calculate the solubility of silver chloride (AgCl) in water at 25◦C. Given that
the Ksp of AgCl is 1.8×10−10.
Solution
Step 1: Write the dissociation of AgCl:
AgCl(s)⇌Ag+(aq) + Cl−(aq)
22
Step 2: Write the expression for the solubility product constant (Ksp):
Ksp = [Ag+][Cl−]
Step 3: Let xbe the solubility of AgCl in moles per liter:
AgCl(s)⇌xmol/L + xmol/L
Step 4: Substitute the solubility values into the Ksp expression:
1.8×10−10 =x×x
x2= 1.8×10−10
x=p1.8×10−10
Step 5: Calculate the solubility of AgCl:
x=p1.8×10−10
x≈1.3×10−5mol/L
Therefore, the solubility of AgCl in water at 25◦C is approximately 1.3×10−5
mol/L.
Question 29
Question
A solution contains 0.1 M lead(II) nitrate (Pb(NO3)2) and 0.1 M sodium sulfate
(Na2SO4). Will precipitation occur if these two solutions are mixed? If so, what
is the chemical equation for the reaction, and what is the concentration of the
remaining ions in solution after the reaction reaches completion?
Solution
Step 1: Determine the possible precipitate by considering the ions in the two
solutions. The possible precipitate is lead(II) sulfate (PbSO4) because Pb2+
ions from lead(II) nitrate and SO2−
4ions from sodium sulfate can react to form
an insoluble lead sulfate precipitate.
Step 2: Write the balanced chemical equation for the precipitation reaction:
Pb(NO3)2+ Na2SO4→PbSO4+ 2NaNO3
Step 3: Determine the concentration of each ion in the solution after the
reaction reaches completion. Initially, the concentration of Pb2+ and SO2−
4ions
is 0.1 M each. After the reaction, all the Pb2+ ions will react with SO2−
4ions to
form PbSO4, which is insoluble and precipitates out of the solution. Therefore,
all Pb2+ ions will be consumed in the reaction, leaving no remaining Pb2+ ions
in solution. The remaining SO2−
4ion concentration will be the excess after all
the Pb2+ ions have reacted.
Therefore, the concentration of remaining SO2−
4ions in the solution is 0.1
M.
23
Question 30
Question
A solution contains 0.1 M of silver nitrate (AgNO3) and 0.2 M of potassium
chloride (KCl). What is the maximum concentration of silver chloride (AgCl)
that can be precipitated from this solution? (Ksp of AgCl is 1.8×10−10)
Solution
Step 1: Write the balanced chemical equation for the precipitation of silver
chloride:
AgNO3+ KCl →AgCl + KNO3
Step 2: Calculate the initial concentration of ions: Given: [Ag] = 0.1 M [Cl]
= 0.2 M
Step 3: Determine the limiting reactant to find the maximum concentration
of silver chloride that can be precipitated: From the balanced chemical equation,
we see that 1 mole of AgCl is formed from 1 mole of Ag and 1 mole of Cl.
Step 4: Calculate the concentration of silver ions and chloride ions formed
by the reaction: The maximum amount of AgCl precipitated will be limited by
the reactant that produces less AgCl. This can be determined by comparing
the initial concentrations of silver ions and chloride ions with the stoichiometry
of the reaction.
Step 5: Calculate the concentration of silver ions remaining: Since Cl is in
excess, all of the AgNO3will be consumed. Therefore, the concentration of
silver ions remaining is 0 M.
Step 6: Calculate the maximum concentration of AgCl that can be pre-
cipitated: Using the concentration of silver ions remaining, we can find the
maximum concentration of AgCl that can be precipitated. This is done by
multiplying the concentration of the limiting ion by the stoichiometry of the
reaction.
Step 7: Calculate the Ksp expression for AgCl: The Ksp expression for AgCl
is given by:
Ksp = [Ag][Cl]
Step 8: Substitute the known values into the Ksp expression and solve for
the maximum concentration of AgCl.
Question 31
Question
Calculate the solubility of silver chloride (AgCl) in water at 25◦C. Given that
the Ksp of AgCl is 1.6×10−10.
24
Solution
Step 1: Write the equation for the dissociation of silver chloride:
AgCl(s)⇌Ag+(aq) + Cl−(aq)
Step 2: Write the expression for the solubility product (Ksp) of AgCl:
Ksp = [Ag+][Cl−]
Step 3: Let the solubility of AgCl be represented as x. Then at equilibrium,
the concentrations of Ag+and Cl−ions will also be x. Thus, substitute these
values into the Ksp expression:
1.6×10−10 =x×x
1.6×10−10 =x2
Step 4: Solve for x:
x=p1.6×10−10 = 1.26 ×10−5
Therefore, the solubility of silver chloride in water at 25◦Cis 1.26 ×10−5
mol/L.
Question 32
Question
Calculate the mass of lead(II) iodide (PbI2) that can be formed from the reaction
of 50.0 mL of 0.150 M lead(II) nitrate (Pb(NO3)2) with excess potassium iodide
(KI). The balanced chemical equation for the reaction is:
Pb(NO3)2(aq) + 2 KI(aq) →PbI2(s) + 2 KNO3(aq)
Solution
Step 1: Write the balanced chemical equation for the reaction.
Pb(NO3)2(aq) + 2 KI(aq) →PbI2(s) + 2 KNO3(aq)
Step 2: Determine the moles of lead(II) nitrate (Pb(NO3)2). Given: Volume
of Pb(NO3)2solution = 50.0 mL = 0.0500 L Molarity of Pb(NO3)2solution =
0.150 M
Moles of Pb(NO3)2= Molarity×Volume = 0.150 mol/L×0.0500 L = 0.00750 mol
Step 3: Use the mole ratio from the balanced equation to determine the
moles of PbI2formed. From the balanced equation, 1 mole of Pb(NO3)2reacts
with 1 mole of PbI2.
Moles of PbI2formed = 0.00750 mol
25
Step 4: Calculate the mass of PbI2formed. The molar mass of PbI2is the
sum of the molar masses of lead (Pb) and iodine (I) in the compound.
Molar mass of P bI2= molar mass of Pb + 2 ×molar mass of I
= 207.2 g/mol + 2 ×126.9 g/mol
= 459.2 g/mol
Mass of PbI2= Moles of PbI2×Molar mass of PbI2= 0.00750 mol×459.2 g/mol
= 3.44 g
Therefore, the mass of lead(II) iodide (PbI2) that can be formed is 3.44
grams.
Question 33
Question
Calculate the mass of precipitate that forms when 50.0 mL of 0.100 M silver
nitrate reacts with excess sodium chloride forming silver chloride, which has a
solubility product constant Ksp = 1.77 ×10−10.
Solution
Step 1: Write the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between silver nitrate (AgNO3)
and sodium chloride (NaCl) to form silver chloride precipitate (AgCl) is:
AgNO3+ NaCl →AgCl + NaNO3
Step 2: Determine the limiting reactant.
Using the balanced chemical equation, we can see that the molar ratio of AgNO3
to AgCl is 1:1. Therefore, all 0.100 M of AgNO3will react with NaCl to form
AgCl.
Step 3: Calculate the moles of AgCl formed.
Since the molar ratio of AgNO3to AgCl is 1:1, the moles of AgCl formed will
be equal to the moles of AgNO3used in the reaction.
Moles of AgCl = Moles of AgNO3= Volume ×Molarity
Moles of AgCl = 0.0500 L ×0.100 mol/L = 0.00500 mol
Step 4: Calculate the mass of AgCl formed.
The molar mass of AgCl is:
Ag: 107.87 g/mol, Cl: 35.45 g/mol
26
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
The mass of AgCl formed is:
Mass of AgCl = Moles of AgCl ×Molar mass of AgCl
Mass of AgCl = 0.00500 mol ×143.32 g/mol = 0.716 g
Therefore, the mass of precipitate (AgCl) that forms is 0.716 grams.
Question 34
Question
Calculate the mass of precipitate that forms when 100.0 mL of a 0.100 M solution
of silver nitrate, AgNO3, is mixed with 150.0 mL of a 0.150 M solution of sodium
chloride, NaCl. Assume the reaction goes to completion, forming solid silver
chloride, AgCl.
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium chloride:
AgNO3+ NaCl →AgCl + NaNO3
Step 2: Determine the limiting reactant. First, calculate the moles of silver
nitrate (AgNO3) and sodium chloride (NaCl) using the provided concentrations
and volumes: For silver nitrate:
moles of AgNO3= M ×V=0.100 mol/L ×0.100 L = 0.0100 mol
For sodium chloride:
moles of NaCl = M ×V = 0.150 mol/L ×0.150 L = 0.0225 mol
Step 3: Determine the limiting reactant. The stoichiometry of the balanced
chemical equation shows that 1 mole of AgNO3reacts with 1 mole of NaCl to
form 1 mole of AgCl. Therefore, the limiting reactant is the one that produces
the least amount of AgCl.
In this case, AgNO3produces only 0.0100 moles of AgCl while NaCl can
produce 0.0225 moles of AgCl. Thus, AgNO3is the limiting reactant.
Step 4: Calculate the mass of precipitate formed. The molar mass of AgCl
is approximately 143.32 g/mol.
Mass of AgCl = moles of AgCl ×molar mass of AgCl
Mass of AgCl = 0.0100 mol ×143.32 g/mol = 1.4330 g
Therefore, the mass of precipitate that forms when 100.0 mL of a 0.100 M
solution of silver nitrate is mixed with 150.0 mL of a 0.150 M solution of sodium
chloride is approximately 1.4330 grams.
27
Question 35
Question
Calculate the solubility of silver chloride (AgCl) in water at 25◦C. The Ksp of
AgCl is 1.8×10−10.
Solution
Step 1: Write the equilibrium equation for the dissolution of silver chloride:
AgCl(s)⇌Ag+(aq) + Cl−(aq)
Step 2: Write the expression for the solubility product, Ksp:
Ksp = [Ag+]×[Cl−]
Step 3: Let xbe the solubility of AgCl, so the equilibrium concentrations
of Ag+and Cl−are both equal to x. Substitute the equilibrium concentrations
into the expression for Ksp:
1.8×10−10 =x×x
x2= 1.8×10−10
x=p1.8×10−10
x= 1.34 ×10−5
Therefore, the solubility of silver chloride in water at 25◦Cis 1.34 ×10−5
mol/L.
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