Download Chemistry 1250 - Sp17 Solutions for Midterm 1

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Chemistry 1250 - Sp17
Solutions for Midterm 1
This material is copyrighted. Any use or reproduction is not allowed except with the expressed written
permission of Dr. Zellmer. If you are taking Chem 1250 you are allowed to print one copy for your own use
during the semester you are taking Chem 1250 with Dr. Zellmer. You are not allowed to disseminate this
material to anyone else during the semester or in the future.
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1) Mixtures must contain two or more different substances.
These substances can be elements (atomic or
molecular) or compounds. There are essentially two types of mixtures, homogeneous and heterogeneous. The
substances in a mixture can be separated into pure substances by physical means. Solutions exist in all phases
(states).
Homogeneous mixtures are uniform throughout and have the same physical and chemical properties
throughout. These are often referred to as solutions. Gases dissolve in each other to form solutions and gases
dissolve in liquids to form solutions.
Heterogeneous mixtures have physical and chemical properties that are NOT uniform throughout the sample.
Heterogeneous mixtures may contain elements and compounds and must contain 2 or more substances.
Pure substances are not mixtures. They are elements or compounds. They can not be separated into simpler
substances through physical means but can be chemically decomposed into simpler pure substances. A pure
substance can contain more than one type of atom. Table salt (NaCl) is a pure compound. Compounds are
always composed of two or more different elements, chemically combined, in fixed proportion by mass. If
only one substance is present the material is a pure substance and the material must be uniform throughout
(homogeneous). Both pure substances and solutions have properties that are uniform throughout.
During a physical change a substance remains that same substance and it’s chemical identity doesn’t change.
Melting point is an intensive prop., it doesn’t depend on the amount of matter present.
E
2)
Physical properties of Bromine (Br):
A. Its density is 3.12 g/cm3.
C. It freezes to form an orange solid.
D. It boils at 58.8EC.
Physical properties depend only on the substance itself and are properties that can be measured without
changing the chemical identity of the substance.
Chemical properties describe how a substance changes during a chemical reaction (i.e. How it reacts and
what it reacts with.)
B
(A, C, D)
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3) Do the calculations in parentheses first.
(341.7 - 22.00) = 319.7 (4 s.f.)
4 s.f. 4 s.f.
(0.00224 x 814005) = 1823.37124 (3 s.f.)
2 s.f.
3 s.f.
341.7
! 22.00
319.7
Multiplication and division rule: answer has same # s.f. as the number with the least # s.f.
Addition and subtraction rule: last place in answer is the same as the last significant place common to all
numbers. Line up the decimals. You can gain or lose s.f. in addition and subtraction.
319.7 + 1823.37124 = 2143.0712 = 2.14 x 103
Can only know the final number to the tens place since the 1823.37124 is only to the tens place. Thus the result
can only be reported to the tens place giving the final answer only 3 s.f..
A
4)
Remember, precision means the degree of reproducibility (how close the measurements are to each other
and how close each measurement is to the average). Accuracy is how close the average is to the true value of
what is measured. To determine precision and accuracy, you should determine the averages of the five trials.
The objects length is 14.54 cm
Student A’s average 14.5
Students B’s average 14.78
Student C’s average 14.66
Student A has done the most accurate work because the average length of 14.5 cm is closest to the true length
of (14.54 cm).
Student B has done the most precise work because the repeated measurements are within 0.2 cm of each other
and 0.12 cm of the average.
A
5)
? km3 = 3.7 × 1020 L × 103 ml × 1 cm3 ×
1L
1 mL
B
(10&2 m)3
(1 cm)3
×
1 km
= 3.7 x 108 km3
(103 m)3
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6)
D
7)
10 3g bicar
1 mg bicar
1 tablet
? tablets = 10.7 kg bicar × --------------- × ---------------- × ------------------------ = 5.1 x 103 tablets
2.1 x 103 mg bicar
1 kg bicar
10!3 g bicar
C
8)
This problem is similar to what was done in exp 1. Exp 1 dealt with taking measurements of mass and volume
of water to determine density. Remember, D = m/V, which can be written as m = D*V (m = D*V + 0). Graphs
were made of mass vs. volume (y vs. x) and the data was fit to a “best-fit” line. A best-fit line averages out the
error in the data. The slope of the line would be equal to the density of the liquid and the line should pass
through the origin, since when there’s zero mass there should be zero volume, and vice versa. Random error
would show up in how well the best-fit line fit the data points (how close the points are to the line). The better
the line fits the points the better the precision (less random, or indeterminate, error). A determinate error such
as using the total mass of the cylinder and liquid, instead of just the mass of the liquid would should up in the yintercept.
Dliq = mlig/Vliq
mlig = Dliq*Vliq
If the mass plotted is the mass of the liquid plus the mass of the cylinder it would show up in the intercept, not
the slope,
mtotal = mcyl + mliq
mliq = Dliq*Vliq
(mtotal ! mcyl) = Dliq/Vliq
mtotal = Dliq/Vliq + mcyl
Thus, the density, slope, is still the density of the liquid and the intercept is the mass of the cylinder.
B
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9)
EF = 9EF (x EC) + 32 EF
5EC
EC = 5EC (xEF - 32 EF)
9EF
K = EC + 273.15
?EF = 9EF (29.85 EC) + 32 EF = 85.73EC = 86E
5EC
? EC = 303 K ! 273.15
= 29.85 EC
= 30 EC
B
10)
The correct statements are 3 and 5:
3) Isotopes of an element differ in the number of neutrons but have the same number of protons and electrons.
5) A neutron is neutral. (The proton has a positive charge and the electron has a negative charge.)
The corrected answers for 1 & 2 would be:
1) The mass number of an atom is the sum of the number of neutrons and protons in the nucleus.
2) Atoms are divisible. (They can be “broken down” into neutrons, protons and electrons.)
4) A proton and neutron have approximately the same mass (1.0073 amu & 1.0088 amu, respectively).
C
(3 & 5 are correct)
11)
The atomic weight (amu) is the weighted average of the masses of the isotopes.
At. wt. = (0.7111) (57.9353) + (0.2889) (59.9332)
41.19779
+
17.31470
At. wt. = 58.51249 = 58.51 amu (4 s.f., final answer can only be reported to the first decimal place)
C
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12)
For ionic compounds you need to know the charge on both the cation and anion and the charges have to balance
since ionic compounds are neutral (like NaCl). For group 1A and 2A metals the cations formed always have a
+1 and +2 charges, respectively. Also, Al, Zn and Ag are always +3, +2 and +1, respectively. Other metals can
have multiple charges depending on the compound and their charges are given as Roman Numerals in
parentheses in the name. In ionic compounds groups 5A, 6A and 7A have charges of !3, !2 and !1
respectively. You have to memorize some polyatomic ions (names, formulas and charges).
For molecular compounds the less electronegative element is generally written first in the formula and is named
first in the name. The second element (more electronegative element) in the formula is named by using the stem
of the name and the suffix -ide. Numerical prefixes, indicating the numbers of each atom, precede the names of
both elements. Common exceptions are for compounds containing H and an element from groups 3A, 4A and
5A (BH3, CH4, NH3, etc.).
How can you tell if the compound is molecular and not ionic? Generally if the compound is composed of a
metal and nonmetal it is ionic. Generally if the compound contains only nonmetals or nonmetal and semimetal
it is molecular. The most common exceptions to this is when a compound contains ammonium ions, NH4+, such
as NH4Cl, (NH4)2SO4, etc. All the elements in the compounds listed (and others with NH4+ ions) contain all
nonmetals but are ionic because of the presence of the NH4+ ions.
The following are the correct formulas for the names given:
a)
iron(III) sulfite
(Ionic)
Charge on iron is +3 (given as Roman Numeral III in name). You should know that SO42! is sulfate with a !2
charge. Sulfite would have 1 fewer O atoms than sulfate and thus is SO32!. Since there are 3 sultites this gives
a total negative charge of !6. The iron has to provide a total positive charge of +6. Since there are 2 iron ions
the charge on each Fe must be +3, as the Roman Numeral in the name indicates.
b)
sulfur hexafluoride
(molecular)
SF6
c)
tetraphosphorus decasulfide
P4S10
***** continued on next page *****
(molecular)
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12)
d)
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(cont.)
aluminum bicarbonate
(Ionic)
Charge on aluminum is +3 (no Roman Numeral is needed since Al is always +3 in compounds). You should
know that CO32! is carbonate with a !2 charge. Adding 1 H+ to carbonate gives bicarbonate (or hydrogen
carbonate), HCO3!. Since there is a +3 charge on the Al and the HCO3! has a !1 charge there has to be 3 HCO3!
ions for every one Al3+. This gives a total positive charge of +3 and a total negative charge of !3.
e)
manganese(IV) iodate
(Ionic)
Charge on manganese is +4 (given as Roman Numeral IV in name). You should know that IO3! is iodate with a
!1 charge. Criss-cross the charges and you get the correct subscripts in this case. The 1 manganese provides a
total positive charge of +4. The 4 iodates gives a total negative charge of !4.
E
13) Scandium chlorate is Sc(ClO3)3
Since the charge on ClO3! is !1 (ClO31!), then the charge on Sc must be +3
carbonate: NO2!
arsenate: SO42!
scandium nitrite: Sc3+ NO2!
Y
Sc(NO2)3
scandium arsenate: Sc3+ SO42!
Y
Sc2(SO4)3
D
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14)
Cl2O7
+
Al(OH)3
6
Al(ClO4)3
+
H2 O
A) Balance Cl (present in greatest number, other than H and O)
3 Cl2O7
+
Al(OH)3
6
2 Al(ClO4)3
+
H2 O
B) Balance Al (present in only one reactant and one product - 2 on right)
3 Cl2O7
+
2 Al(OH)3
6
2 Al(ClO4)3
+
H2 O
C) Balance H (present in only one reactant and one product - 6 on left; need 6 on right)
3 Cl2O7
+
2 Al(OH)3
6
2 Al(ClO4)3 +
3 H2 O
The sum of the coefficients of the products = 3 + 2 = 5
C
15)
PtCl4
+
XeF2
6
PtF6
+
ClF
+
Xe
A) Balance Cl (While F appears in greatest amount the Cl is in only 1 cmpd on each side)
PtCl4
+
6
XeF2
PtF6
+
4 ClF
+
Xe
B) Balance F (Pt is already bal., leave Xe until end since it is in elemental form on right)
PtCl4
+
5 XeF2
6
PtF6
+
4 ClF
+
Xe
C) Balance Xe (Balance the 5 Xe on left with 5 Xe on right)
1 PtCl4
+
5 XeF2
6
1 PtF6
+
4 ClF
+
5 Xe
(coefficients of 1's in front of the compounds are usually not shown)
The sum of the coefficients of the reactants AND products = 1 + 5 + 1 + 4 + 5 = 16
D
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16) NH3
+
9
6
O2
NO
+
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H2 O
A) Balance H (Usually left until end but N is already balanced & O in elemental form.)
+
2 NH3
O2
6
NO
O2
6
2 NO
+
3 H2 O
B) Balance N
2 NH3
+
+
3 H2 O
Need 5 O atoms from O2
C) Balance O
6
2&0+3&0=
(5 & 0 atoms)
Use 5/2 O2 ( ½ of 5 O = O gives 5 O atoms)
2 NH3
+
O2
6
2 NO
5 O2
6
4 NO
(5/2)
+
3 H2 O
D) Multiply by 2:
4 NH3
+
+
6 H2 O
The sum of the coefficients of the reactants & products = 4 + 5 + 4 + 6 = 19
C
17) Need molecular weight for Pt(NH3)2Cl2 to determine weight % of N
MW
A
= 1(AWPt)
= 1(195.1)
= 300.068
+
+
2(AWN)
2(14.01)
+
+
6(AWH)
6(1.008)
+
+
2(AWCl)
2(35.45)
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18)
Have two molecules AZ(g) and A2Z(R) composed of atoms A and Z (the states of the two molecules is
immaterial). There are 6.00 g of each. Based on the formulas the A2Z molecule has a larger MW
(molar mass). The molar mass of a substance is inversely proportional to the moles (MM = g/mol) and
thus the moles is inversely proportional to the MM,
MM = g/mol
mol = g/MM
Thus, since the mass of each cmpd is the same (6.00 g) the substance with the smaller molar mass will
have the larger number of moles and thus the larger number of molecules (since molecules and moles
are directly proportional). AZ has a smaller molar mass than A2Z and thus AZ has the larger number of
moles and molecules.
C
19)
E
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20)
This is an empirical formula problem using combustion analysis. In CA the sample is combusted. All
the Carbon winds up in the CO2 and the Hydrogen winds up in the H2O . If there is one other atom its
mass can be determined by remembering the conservation of mass, in this case oxygen.
In this problem a 0.162 g (162 mg) sample produces 0.440 g (440 mg) of CO2 (assume 3 s.f. since
others have 3) & 0.126 g (126 mg) of H2O.
Determine moles and mass of C and H in the sample:
1 mol CO2
1 mol C
? mol C = 0.440 g CO2 × ----------------- × --------------- = 0.0099977 mol C (3 s.f.)
44.01 g CO2
1 mol CO2
1 mol H2O
2 mol H
? mol H = 0.126 g H2O × ------------------ × ---------------- = 0.01398 mol H (3 s.f.)
18.016 g H2O
1 mol H2O
Need moles of oxygen but have to get mass first. Need to calculate mass of C and H and subtract from
the total mass of sample.
12.01 g C
? g C = 0.009997 mol C × --------------- = 0.12007 g C (3 s.f.)
1 mol C
1.008 g H
? g H = 0.01398 mol H × ----------------- = 0.01409 g H (3 s.f.)
1 mol H
? g N = 0.162 g sample ! (0.12007 g C + 0.01409 g H) = 0.02783 g N
Find moles of N:
1 mol N
? mol N = 0.02783 g N × -------------- = 0.001986 mol N (3 s.f.)
14.01 g N
***** continued on next page *****
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20) (cont.)
Empirical Formula calculations:
Divide each of the moles by the smallest number of moles (in this case N).
0.0099977 mol C
C: ------------------------ = 5.03 = 5
0.001986 mol N
0.01398 mol H
H: ----------------------- = 7.04 = 7
0.001986 mol N
0.001986 mol N
N: ------------------------ = 1.00 = 1
0.001986 mol N
C5H7N
A
21)
All soluble ionic substances (salts) are strong electrolytes. All the substances are soluble and come apart
to form ions in solution. Use the subscripts in the formulas for each element (stoichiometry) to
determine the conc. of each ion in solution.
D
Na+
+ I!
0.060 M
0.060 M
a) NaI
0.060 M
—>
b) Cu(IO2)2
0.030 M
—>
Cu2+
+
2 IO2!
0.030 M
2(0.030 M) = 0.060 M
c) FeI3
0.025 M
—>
Fe3+
+
I!
0.025 M
3(0.025 M) = 0.075 M
d) BaI2
0.040 M
—>
Ba2+
+
I!
0.040 M
2(0.040 M) = 0.080 M
e) KIO3
0.090 M
—>
K+
+
IO3!
0.090 M
0.090 M
This iodite NOT iodide
This iodate NOT iodide
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22)
This is a limiting reactant problem. These are just stoichiometry problems. There is more than one way
to do a LR problem. In this case it’s asking for the limiting reactant and the mass of excess reactant
remaining after completion of the reaction.
2 Fe(OH)3 (s) + 3 H2SO4 (aq) v Fe2(SO4)3 (aq) + 6 H2O (R)
1) Method 1: Calculate which reactant gives the smallest number of moles of product, and then use to
determine how much of the excess reactant would be used (and remains).
Calculate mol Fe2(SO4)3 from Fe(OH)3 and H2SO4 :
1 mol Fe2(SO4)3
? mol Fe2(SO4)3 = 1.60 mol Fe(OH)3 × ---------------------- = 0.80 mol Fe2(SO4)3 (2 s.f.)
2 mol Fe(OH)3
1 mol Fe2(SO4)3
? mol Fe2(SO4)3 = 2.00 mol H2SO4 × ---------------------- = 0.67 mol Fe2(SO4)3 (2 s.f.) LR
3 mol H2SO4
Fewer moles of Fe2(SO4)3 obtained from the H2SO4 so the H2SO4 is the LR and only 0.67 mol of
Fe2(SO4)3 are formed. Calculate the moles of Fe(OH)3 required to produce 0.67 mol of Fe2(SO4)3
2 mol Fe(OH)3
? mol Fe(OH)3 = 0.67 mol Fe2(SO4)3 × --------------------- = 1.33 mol Fe(OH)3 used (2 s.f.)
used
1 mol Fe2(SO4)3
? mol Fe(OH)3 = 1.60 Fe(OH)3 ! 1.33 mol Fe(OH)3 = 0.27 mol Fe(OH)3 remaining
H2SO4 is LR
0.67 mol Fe2(SO4)3 produced
0.27 mol Fe(OH)3 remaining
2) Method 2: Determine mole ratio of the two reactants & compare it to the mole ratio in the bal. eqn.
2.00 mol H2SO4
3 mol H2SO4
-------------------------- < -----------------------2 mol Fe(OH)3
1.60 mol Fe(OH)3
Since the actual ratio of H2SO4 to Fe(OH)3 (ratio = 1.25) is less than that in the bal. eqn. (ratio = 1.5) this
means there isn’t enough H2SO4 so it is the LR. Can then calculate the moles of Fe2(SO4)3 produced
from the H2SO4 and the
1 mol Fe2(SO4)3
? mol Fe2(SO4)3 = 2.00 mol H2SO4 × ---------------------- = 0.67 mol Fe2(SO4)3 (2 s.f.)
3 mol H2SO4
2 mol Fe(OH)3
? mol Fe(OH)3 = 2.00 mol H2SO4 × --------------------- = 1.33 mol Fe(OH)3 used (2 s.f.)
3 mol H2SO4
(0.27 mol remains)
B
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23)
14
1 and 4 produce gas
1) BaSO3 (s) + 2 HCl (aq)
ionic
2) Fe (s)
+
6
ionic
element
element
6
FeCl2 (aq)
+
with gas
gas
Cu (s)
single repl.
element
2 Li3N (s)
combination
(2 reactants 6 1 product)
1 compound
2 H2O (R)
6
water
2 KOH (aq)
+
ionic
5) CaSO4 (aq) + (NH4)2S (aq)
ionic
water
ionic
element
+
BaCl2 (aq) + H2O (R) + SO2 (g) double repl.
ionic
CuCl2 (aq)
3) 6 Li (s) + N2 (g)
4) 2 K (s)
6
acid
element
E
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6
ionic
H2 (g)
element
CaS (s) + (NH4)2SO4 (aq)
ionic
ionic
single repl.
with gas
double repl.
with ppt.
(1 & 4)
24)
See above answers. 2 and 4 are single-replacement (displacement) reactions.
2) Fe (s)
+
element
4) 2 K (s)
element
D
(2, 4)
CuCl2 (aq)
6
ionic
+
2 H2O (R)
water
FeCl2 (aq)
+
Cu (s)
ionic
6
2 KOH (aq)
ionic
single repl.
element
+
H2 (g)
element
single repl.
with gas
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15
25)This is a double-replacement reaction.
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Write the ionic equation and cancel out everything that appears the
same way on both sides of the equation.
MgCl2 (aq) + Na3PO4 (aq)
6 Mg2+ (PO4)3! (s) + Na1+ Cl1! (aq)
3 MgCl2 (aq) + 2 Na3PO4 (aq)
6 Mg3(PO4)2 (s) + 6 NaCl (aq)
3 Mg2+ (aq) + 6 Cl& (aq) + 6 Na+(aq) + 2 PO43! (aq)
3 Mg2+ (aq) + 2 PO43! (aq)
6 Mg3(PO4)2 (s) + 6 Na+(aq) + 6 Cl& (aq)
6 Mg3(PO4)2 (s)
C
26)
This is a solution stoichiometry problem an exchange reaction. Write the balanced equation.
2 LaCl3 + 3 Na2C2O4
v
2 La2(C2O4)3 + 3 NaCl
This is like a gram-to-gram stoichiometry problem but instead of first converting grams to moles using
molar mass you need to use volume and molarity. Remember, molarity (mol/L) is a conversion between
moles of solute and volume of solution.
mol solute
M = ----------------L soln
Determine the vol. of 0.00927 M LaCl3 required to react with 13.95 mL of 0.0225 M Na2C2O4.
1 L soln
0.0225 mol Na2C2O4
2 mol LaCl3
1 L LaCl3
? mL LaCl3 = 13.95 mL Na2C2O4 x ---------------- x -------------------------- x -------------------- x -----------------------soln.
103 mL soln
1 L soln
3 mol Na2C2O4
0.00927 mol LaCl3
= 0.022572 L LaCl3 soln
= 22.6 mL LaCl3 soln (3 s.f.)
B
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27)
Myocrisin (Na2C4H3O4SAu) is used to treat rheumatoid arthritis. A patient receives an injection from a
solution that contains 50.0 mg of Myocrisin, in 0.500 mL. What is the molarity of myocrisin?
Let MY = Myocrisin
mol solute
M = ----------------L soln
Molarity
50.0 mg MY
103 mL soln
1 g MY
1 mol MY
? mol/L MY = ---------------- × ------------------ × -----------×
-----------------0.500 mL
1 L soln
103 mg
390.10 g MY
= 2.56 x 10!1 mol MY/L soln
= 2.56 x 10!1 M MY
A
28)
a) HClO3
(+1)H+XCl+3(&2)O = 0
b) Cl2
c) Cl2O7
XCl = 0
d) CaCl2
2XCl+7(-2)O = 0
2XCl&14 = 0
XCl = +5
(+2)Ca+2XCl = 0
e) ClO2&
XCl+2(-2)O = &1
XCl = &1
XCl = +3
&1
+3
XCl = +7
+5
0
+7
&1,
0,
CaCl2, Cl2,
d
E
b
+3,
+5,
ClO2&, HClO3,
e
a
+7
Cl2O7
c
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29)
The reactions are redox (oxidation-reduction) reactions. To be a redox (oxidation-reduction) reaction the
oxidation numbers must change. More specifically the reactions are all displacement (single replacement)
reactions. You need to figure out what’s being oxidized and reduced. Then you use the activity series.
Remember, ease of oxidation increases from bottom to top in the series. The more easily oxidized
substance is higher in the table.
Remember (OIL RIG):
oxidation:
loss of e-, inc. in oxidation #
reduction:
gain of e-, dec. in oxidation #
Reaction (a):
Fe(s) + 2 HI(aq) 6 H2(g) + FeI2(aq)
0
+1
0
Fe oxidized
H reduced
+2
In the activity series table iron is higher than hydrogen meaning iron is more easily oxidized.
Since Fe is being oxidized this reaction will occur as written going left to right. Thus the Fe
replaces H+ in solution.
Reaction (b):
2 Co(s) + 2 AgCl(s) 6 2 Ag(s) + 2 CoCl2 (aq)
0
+1
0
+2
Co oxidized
Ag reduced
In the activity series table cobalt is higher than silver meaning cobalt is more easily oxidized.
Since Co is being oxidized this reaction will occur as written going left to right. Thus the Co
replaces Ag+ in solution.
Reaction (c):
2 Al(s) + 3 Pb(NO3)2(aq) 6 3 Pb(s) + 2 Al(NO3)3 (aq)
0
+2
0
+3
Al oxidized
Pb reduced
In the activity series table aluminum is higher than lead meaning aluminum is more easily
oxidized. Since Al is being oxidized this reaction will occur as written going left to right. Thus
the Al replaces Pb2+ in solution.
***** continued on next page *****
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29)
18
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(cont.)
Reaction (d):
Proceeding in this way you find reaction (d) can’t occur as written:
Mg(s) + Ba(OH)2(aq) 6 Ba(s) + Mg(OH)2 (aq)
0
+2
0
+2
Mg oxidized
Ba reduced
However, from the table Ba is more easily oxidized than Mg. This reaction won’t occur
spontaneously left to right but instead the reverse reaction would be spont.
Reaction (e):
Zn(s) + Pb(NO3)2(aq) 6 Pb(s) + Zn(NO3)2 (aq)
0
+2
0
+2
Zn oxidized
Pb reduced
In the activity series table zinc is higher than lead meaning zinc is more easily oxidized. Since
Zn is being oxidized this reaction will occur as written going left to right. Thus the Zn replaces
Pb2+ in solution.
D
30)
N: reduced
HNO3: oxidizing agent
E
(1, 2 & 5 are Correct)
Se: oxidized
Se: reducing agent