Download Aqueous Reactions and Solution Stoichiometry (Chapter 4) Past

Aqueous Reactions and Solution Stoichiometry (Chapter 4)
Past Quizzes and Tests
My answers are in bold-faced underlined italics.
1.
2.
3.
4.
How many moles of Ca(NO3)2 are contained in 150.0-mL of a 0.245-M solution of
Ca(NO3)2?
150.0mL 0.245molCa ( NO3 ) 2
? molCa ( NO3 ) 2 =
x
= 0.03675molCa ( NO3 ) 2
1
1000mL
How many grams of KCl are required to make 450.0-mL of a solution that is
0.100-M KCl?
450.0mL 0.100molKCl 74.5KCl
? gKCl =
x
x
= 3.353gKCl
1
1000mL
1molKCl
34.5-g of BaBr2 are dissolved in water to make a solution that has a total volume of
750.0-mL. What is the molarity of BaBr2, Ba2+, and Br- in the solution?
34.5 gBaBr2
# molsolute
297 g / mol
M =
=
= 0.155MBaBr2
# LSolution
0.750 L
For the molarity of the other species, realize that each mol of BaBr2 provides one mol
of Ba2+ and 2 mol of Br-. Their concentrations are 1x0.155M = 0.155 M Ba2+ and 2 x
0.155 M = 0.310 M Br-.
0.415-g of potassium hydrogen phthalate are dissolved in water and phenolphthalein
indicator added. The equivalence point is reached with the addition of 31.95-mL of a
NaOH solution. What is the molarity of the NaOH solution? (Molar mass of KHP =
204.1 g/mol)
Potassium hydrogen phthalate, KHC8H4O4, reacts with NaOH according to the
equation:
KHC8H4O4 (aq) + NaOH (aq) 6 NaKC8H4O4 (aq) + HOH (R)
From the equation, note that one mole of KHC8H4O4 reacts with 1 mol of NaOH. So, if we
know how many mol of KHC8H4O4 were in the flask initially, we can determine the number
of moles of NaOH in the volume of NaOH delivered.
# molKHC8 H 4O4 =
0.415 gKHC8 H 4O4
= 2.03x10 −3 molKHC8 H 4O4
204.1g / mol
so # molNaOH = 2.03x10 −3 molNaOH
so M NaOH
# molNaOH 2.03x10 −3 molNaOH
=
=
= 6.36 x10 −2 MNaOH
# LNaOH
0.03195Lsolution
5.
6.
Find the oxidation number of the underlined element in each of the following species.
AsO43- +5
Ba(NO2)2 +3
NH4+
HCO3 G
-3
N2O5 +5
+4
Complete and balance the following equations. Also write the total ionic and net ionic
equations for each.
3 SrCl2 (aq) + 2 Na3PO4 (aq) ------> Sr3(PO4)2 (s) + 6 NaCl (aq)
Total Ionic:
3 Sr2+ (aq) + 6 Cl- (aq) + 6 Na+ (aq) + 2 PO43- (aq) Æ Sr3(PO4)2 (s) + 6Na+ (aq) + 6 Cl- (aq)
Net Ionic:
3 Sr2+ (aq) + 2 PO43- (aq) Æ Sr3(PO4)2 (s)
3 Li (s) + FeCl3 (aq) ----> 3 LiCl (aq) + Fe (s)
Total Ionic:
3 Li (s) + Fe3+ (aq) + 3 Cl- (aq) Æ 3 Li+ (aq) + 3 Cl- (aq) + Fe (s)
Net Ionic:
7.
3 Li (s) + Fe3+ (aq) Æ 3 Li+ (aq) + Fe (s)
Classify each of the following as to whether it is a salt, acid, or base (S/A/B); soluble or
insoluble (S/I); and strong or weak/non electrolyte (S/W).
Compound
A/B/S
S/I
S/W
Compound
A/B/S
S/I
S/W
HNO2
A
S
W
PbBr2
S
I
S
BaSO4
S
I
S
Cu(NO3)2
S
S
S
HC2H3O2
A
S
W
Fe2O3
S
I
S
H2C8H4O4
A
S
W
(NH4)2S
S
S
S
Fe(OH)3
B
I
S
H2S
A
S
W
8.
Complete the following metathesis reaction and write the total ionic equation and the net
ionic equation.
Molecular equation:
CaCl2 (aq) +
Pb(NO3)2 (aq)
-----> Ca(NO3)2 (aq) + PbCl2 (s)
Total Ionic:
Ca2+ (aq) + 2 Cl- (aq) + Pb2+ (aq) + 2 NO3- (aq) Æ Ca2+ (aq) + 2 NO3- (aq) + PbCl2 (s)
Net Ionic:
9.
2 Cl- (aq) + Pb2+ (aq)
Æ PbCl2 (s)
Complete and balance the following metathesis reactions. Further write the total ionic
and net ionic equations for each:
3 BaCl2 (aq)
+ 2 Na3PO4 (aq) Æ Ba3(PO4)2 (s) + 6 NaCl (aq)
Total Ionic:
3 Ba2+(aq) + 6 Cl-(aq) + 6 Na+ (aq) + 2 PO43- (aq) Æ Ba3(PO4)2 (s) + 6 Na+ (aq) + 6 Cl-(aq)
Net Ionic:
3 Ba2+(aq) + 2 PO43- (aq) Æ Ba3(PO4)2 (s)
HC2H3O2 (aq)
+
KOH (aq) Æ HOH (R) + K C2H3O2 (aq)
Total Ionic:
HC2H3O2 (aq) + K+ (aq) + OH- (aq) Æ HOH (R) + K+(aq) + C2H3O2- (aq)
Net Ionic:
10.
HC2H3O2 (aq) + OH- (aq) Æ HOH (R) + C2H3O2- (aq)
Classify each of the following solutes as acids, bases, or salts. For each indicate whether
it is a strong or weak/non electrolyte (SE or WE) and whether the substance is soluble (S)
or insoluble (I) in aqueous solution.
H2SO3 Acid, weak, soluble
Ba3(PO4)2 Salt, Strong, Insoluble
NaCl
Salt, Strong, Soluble
NH4Cl Salt, Strong, Soluble
BaBr2 Salt, Strong, Soluble
11.
For each of the following:
a. State whether it is an acid(A), base(B), or salt(S).
b. State whether it is a strong electrolyte (SE) or weak/nonelectrolyte(WE)
c. State whether it is soluble (S) or insoluble (I).
Substance
Acid(A)/Base(S)/Salt(S)
Strong Electrolyte (SE)/Weak Electrolyte (WE)
Soluble (S)/Insoluble (I)
KCl
S
SE
S
(NH4)2S
S
SE
S
HNO2
A
WE
S
PbCl2
S
SE
I
Fe(OH)3
B
SE
I
Ba3(PO4)2
S
SE
I
HClO4
A
SE
S
HBrO
A
WE
S
Ba(C2H3O2)2
S
SE
S
Mg(OH)2
B
SE
I
12.
Classify each of the following reactions as to whether it is combustion, decomposition,
combination, metathesis, or oxidation-reduction. Note some equations may fall into more
than one category. State all categories that apply in each case.
Equation
2 NaHCO3 (s) ----> Na2CO3 (s) + H2O (R) + CO2 (g)
BaO (s) + H2O (R) ----> Ba(OH)2 (aq)
Ba(NO3)2 (aq) + Na2SO4 (aq) ----> BaSO4 (s) + 2 NaNO3 (aq)
Cu (s) +
2 AgNO3 (aq) ----> 2 Ag (s) + Cu(NO3)2 (aq)
CH4 (g) +
2 O2 (g) —> CO2 (g) + 2 H2O (R)
Classification
decomposition
combination
metathesis
oxidation-reduction
combustion, oxidationreduction
13.
Complete and balance each of the methathesis equations below and write the total ionic and
net ionic equation for each.
H2SO4 (aq) +
Complete equation: H2SO4 (aq) +
2 NaOH (aq)
2 NaOH (aq)
----->
----->
HOH (ℓ) + Na2SO4 (aq)
Total ionic equation:
2 H+ (aq) + SO42- (aq) + 2 Na+ (aq) + 2 OH- (aq) Æ HOH (ℓ) + 2 Na+ (aq) + SO42- (aq)
Net ionic equation:
2 H+ (aq) + 2 OH- (aq) Æ HOH (ℓ)
BaCl2 (aq) +
Pb(NO3)2 Æ
Complete equation:
Pb(NO3)2 Æ
BaCl2 (aq) +
Ba(NO3)2 (aq) + PbCl2 (s)
Total ionic equation:
Ba2+ (aq) + 2 Cl- (aq) + Pb2+ (aq) + 2 NO3- (aq) Æ Ba2+ (aq) + 2 NO3- (aq) + PbCl2 (s)
Net ionic equation:
14.
a.
2 Cl- (aq) + Pb2+ (aq) Æ PbCl2 (s)
In the standardization reaction of a solution of NaOH, 0.4529-g of potassium hydrogen
phthalate (molar mass = 204.2) is weighed into an Erlenmeyer flask. Water and
phenolphthalein are added to the flask to dissolve the phthalate and provide an
indicator for the titration. 34.75-mL of the NaOH solution were required to reach the
endpoint in the titration between the potassium hydrogen phthalate and the NaOH.
What is the molarity of the NaOH solution?
# molKHP =
0.4529 gKHP
= 0.002218molKHP
204.2 g
# molKHP = # molNaOH = 0.002218molNaOH
MNaOH =
b.
0.002218molNaOH
= 0.06383MNaOH
0.03475L
The same NaOH solution from part a is used to titrate 25.00-mL of a H2SO4 solution.
The titration requires 29.45-mL of the NaOH solution. What is the molarity of the
H2SO4?
# molNaOH = 0.02945 LNaOHSolution
x 0.06383MNaOH = 0.001880molNaOH
1
x # molNaOH = 0.0009399
2
0.0009399molH 2 SO4
MH 2 SO4 =
= 0.03760 M
0.025Lsolution
#molH 2 SO4 =
15.
Indicate for each of the following aqueous solutions whether there would be a precipitate
formed upon mixing or not. If a precipitate is formed indicate the formula and name for the
precipitate. (Hint: Think metathesis)
a.
b.
c.
d.
e.
16.
Pb(NO3)2 and KCl Yes; PbCl2; lead (II) chloride
BaBr2 and NH4NO3 No; both Ba(NO3)2 and NH4Br are soluble
AgNO3 and CaCl2 Yes; AgCl; silver chloride
BaCl2 C2H2O and Na3PO4 C12 H2O Yes; Ba3(PO4)2; barium phosphate
KBr and NaCl No; both KCl and NaBr are soluble
For one of the reactions that forms a precipitate in question 15:
a. Write the balanced chemical equation.
b. Write the total ionic equation.
c. Write the net ionic equation.
For reaction a of problem 15:
Total molecular equation: Pb(NO3)2 (aq) + 2 KCl (aq) Æ
PbCl2 (s) + 2 KNO3 (aq)
Total ionic equation:
Pb2+ (aq) + 2 NO3- (aq) + 2 K+(aq) + 2 Cl- (aq) Æ PbCl2 (s) + 2 K+ (aq) + 2 NO3- (aq)
Net ionic equation:
17.
Pb2+ (aq) + 2 Cl- (aq) Æ PbCl2 (s)
Balance each of the following equations and classify it as combustion, decomposition,
combination, methathesis, and/or oxidation-reduction. Note that some equations may fall into
more than one category. State all categories that apply in each case.
Equation
2 C(gr) + ___ O2 (g) Æ 2 CO (g)
Classification
combustion and redox
___ NH3 (g) + ___ HCl (g) Æ ___ NH4Cl (s)
combination
___ Cr(s) + ___ NiSO4 (aq) Æ ___ Ni (s) + ___ CrSO4 (aq)
redox
___ HC2H3O2 (aq) + ___NaOH(aq) Æ ___ HOH (R) + ___ NaC2H3O2 (aq) metathesis
2 H2O (R) Æ 2 H2 (g) + ___ O2 (g)
18.
_decomposition/redox
Classify each of the following as an acid (A), base (B), or salt (S). Further classify
each as to whether it is a strong electrolyte (SE) or weak/non electrolyte (WE).
HBr
Acid,SE
Cu(OH)2
Base, SE
BaSO4 Salt,SE
Mn(NO3)2 Salt, SE
NaOH
Base, SE
K2C2O4
NH4Cl
Salt,SE
NaF
Salt, SE
AgCl
Salt,SE
BaBr2
Salt, SE
Salt, SE
19.
Complete and balance each of the following metathesis reactions. Further write the
total ionic and net ionic equation for each.
a.
Molecular:
Ba(NO3)2 (aq) +
Na2SO4 (aq) Æ BaSO4 (s) + 2 NaNO3 (aq)
Total Ionic:
Ba2+ (aq) + 2 NO3- (aq) + 2 Na+ (aq) + SO42- (aq) Æ BaSO4 (s) + 2 Na+ (aq) + 2 NO3- (aq)
Net Ionic:
b.
Molecular:
Ba2+ (aq) + SO42- (aq) Æ BaSO4 (s)
2 HCl (aq) + Ba(OH)2 (aq) Æ 2HOH (ℓ) + BaCl2 (aq)
Total Ionic:
2H (aq) + 2Cl- (aq) + Ba2+ (aq) + 2 OH- (aq) Æ 2HOH (ℓ) + Ba2+ (aq) + 2 Cl- (aq)
+
Net Ionic:
20.
2H+ (aq) + 2 OH- (aq) Æ 2HOH (ℓ)
Or
+
H (aq) + OH (aq) Æ HOH (ℓ)
30.00-mL of a 0.130-M solution of AgNO3 is mixed with 20.00-mL of a 0.120-M
solution of BaCl2. How many grams of AgCl could be formed in this process?
Since we are given amounts of two reactants, we need to figure out which one is the
limiting reactant The problem involves determining which reactant will be used up. The
determination of the limiting reactant requires determining how many moles of each
reactant are present in the mixture. This is done by multiplying the number of liters by the
molarity of the respective solution.
# molAgNO3 = 0.0300 Lx0.130 MAgNO3 = 0.003900molAgNO3
# molBaCl2 = 0.0200 Lx0.120 M = 0.002400molBaCl2
0.003900 mol AgNO3 0.002400 mol BaCl2
2 AgNO3 (aq) + BaCl2 (aq) Æ
2 AgCl (s)
2 mol AgNO3
2 x 169 g AgNO3
1 mol BaCl2
1 x 208 g BaCl2
2 mol AgCl
2 x 142.5 g AgCl
+
Ba(NO3)2 (aq)
1 mol Ba(NO3)2
1 x 261 g Ba(NO3)2
To check for limiting reactant, compare the two ratios:
0.003900molAgNO3
0.002400molBaCl2
= 0.001950
= 0.002400
2molAgNO3
1molBaCl2
Since the AgNO3 ratio is lower, it is the limiting reactant. To find the grams of AgCl, use the
AgNO3 since it is limiting.
# gAgCl =
0.0039molAgNO3 2 x142.5 gAgCl
x
= 0.5558gAgCl
1
2molAgNO3
21.
Identify each of the following as an acid, base, or salt. Further identify each as
a strong electrolyte (SE) or weak/non electrolyte (WE).
Ba(NO3)2 Salt,SE
22.
LiOH
Base,SE
HNO2
Acid,WE
HF Acid,WE
Ca(C2H3O2)2 Salt,SE
Label each of the following as an acid, base, or salt. Further identify each as a
strong electrolyte (SE), or a weak/non electrolyte (WE).
Ba(NO3)2 Salt;SE
CaBr2 Salt;SE
HC2H3O2 Acid;WE
HF
LiOH
Base;SE
HOH A/B;WE
HNO2
Acid;WE
Cu(OH)2 Base;SE
RbC8H14O2 Salt;SE
23.
Acid;WE
Ba3(PO4)2 Salt;SE
25.00-mL of 0.10 M HCl are mixed with 35.00 mL of 0.15 M Pb(NO3)2.
a. Write a balanced chemical equation for this metathesis reaction.
2 HCl (aq) + Pb(NO3)2 (aq) Æ
2 HNO3 (aq) +
PbCl2 (s)
b. Which is the limiting reactant?
Check number of moles of each:
# mol HCl = 0.025L x 0.20 M HCl = 0.0050 mol HCl
# mol Pb(NO3)2 = 0.035 L x 0.15 M Pb(NO3)2 = 0.00525 mol Pb(NO3)2
Need twice as many mol of HCl as Pb(NO3)2, meaning we would need 2 x 0.00525 mol
Pb(NO3)2 which is 0.01050 mol of the HCl. There is not enough. HCl is limiting.
c. How many grams of PbCl2 could be formed in the reaction?
Use the HCl since it is limiting.
# molPbCl2 =
0.0050molHCl 1x 278 gPbCl2
x
= 0.695 gPbCl2
1
2molHCl
24.
Balance each of the following equations. Also classify each as to whether it is
combination, decomposition, single displacement, metathesis, or redox. Some
equations may fall into more than one category. Be sure to list all categories
that apply for each equation.
Classification(s)
2SO3 (g)
6
2SO2 (g) + ___ O2 (g)
___ BI3 + ___ GaF3
___ P4
+ 5 O2
2 C4H10 + 13 O2
25.
6
decomposition,redox
___ BF3 + ___ GaI3
6
___ P4O10
6
8 CO2 + 10 H2O
metathesis
combination,redox
combustion,redox
___ Zn + 2 HCl 6 ___ ZnCl2 + ___ H2
displacement,redox
Complete and balance the following metathesis reactions. Further write the total
ionic and net ionic equation for each.
BaCO3 (s) +
6 Ba(NO3)2 (aq)
2 HNO3 (aq)
+ H2CO3 (aq)
Total Ionic:
BaCO3 (s) + 2 H+ (aq) + 2 NO3- (aq) Æ Ba2+ (aq) + 2 NO3- (aq) + H2CO3 (aq)
Net Ionic:
BaCO3 (s) + 2 H+ (aq) Æ Ba2+ (aq) + H2CO3 (aq)
(Note: In water, the H2CO3 breaks into water and carbon dioxide – you would see bubbles coming
off)
H2SO4 (aq)
+
2 NaOH (aq)
6
Na2SO4 (aq) + 2 HOH (ℓ)
Total Ionic:
2 H+ (aq) + SO42- (aq) + 2 Na+ (aq) + 2 OH- (aq) Æ 2 Na2+ (aq) + SO42- (aq) + 2 HOH (ℓ)
Net Ionic:
2 H+ (aq) + 2 OH- (aq) Æ 2 HOH (ℓ)
Or
H+ (aq) + OH- (aq) Æ
HOH (ℓ)
26.
125.0-mL of 0.183 M HNO3 is diluted with water to 500.0-mL in a volumetric
flask. What is the new concentration of the HNO3?
M 1V1 = M 2V2
0.183Mx125.0mL = M 2 x500.0mL
M2 =
0.183Mx125.0mL
= 0.458M
500.0mL