CHAPTER 3
CH7 PROBLEM SOLVING CLASS
R.D. A. BOLINAS
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3.10 Calculate the molar mass of
each of the following:
(a) N2O4
(b) C8H10
(c) MgSO4 ⦁7H2O
(d) Ca(C2H3O2)2
• a) N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol
•
b) C8H10 = 8(12.01) + 10(1.008) = 106.16 g/mol
•
c) MgSO4•7H2O = 24.31 + 32.07 + 11(16.00) +
14(1.008) = 246.49 g/mol
•
d) Ca(C2H3O2)2 = 40.08 + 4(12.01) + 6(1.008) +
4(16.00) = 158.17 g/mol
3.14 Calculate each of the following
quantities:
(a) Total number of ions in 38.1 g of
SrF2
(b) Mass in kilograms of 3.58 mol of
CuCl2 2H2O
(c) Mass in milligrams of 2.88 x1022
formula units of Bi(NO3)3 ⦁ 5H2O
3.18 Calculate each of the following:
(a) Mass % of I in strontium periodate
(b) Mass % of Mn in potassium
permanganate
3.23 Hemoglobin, a protein in red blood
cells, carries O2 from the lungs to the
body’s cells. Iron (as ferrous ion, Fe2+ )
makes up 0.33 mass % of hemoglobin.
If the molar mass of hemoglobin is
6.8x 104 g/mol, how many Fe2+ ions are in
one molecule?
3.26 What is the empirical formula
and empirical formula mass
for each of the following compounds?
(a) C4H8
(b) C3H6O3
(c) P4O10
(d) Ga2(SO4)3
(e) Al2Br6
3.32 A 0.370-mol sample of a metal oxide
(M2O3) weighs 55.4 g.
(a) How many moles of O are in the
sample?
(b) How many grams of M are in the
sample?
(c) What element is represented by the
symbol M?
3.38Write balanced equations for each of the following
by inserting the correct coefficients in the blanks:
(a) __Cu(NO3)2(aq) + __KOH(aq) ⟶
__Cu(OH)2(s) + __KNO3(aq)
(b) __BCl3(g) + __H2O(l) ⟶__H3BO3(s) + __HCl(g)
(c) __CaSiO3(s) + __HF(g) ⟶
__SiF4(g) + __CaF2(s) + __H2O(l)
(d) __(CN)2(g) __H2O(l) ⟶ __H2C2O4(aq) __NH3(g)
• a) Cu(NO3)2(aq) + 2 KOH(aq) 
Cu(OH)2(s) + 2 KNO3(aq)
• b) BCl3(g) + 3 H2O(l)  H3BO3(s) + 3 HCl(g)
• c) CaSiO3(s) + 6 HF(g)  SiF4(g) + CaF2(s) + 3 H2O(l)
• d) (CN)2(g) + 4 H2O(l)  H2C2O4(aq) + 2 NH3(g)
3.41 The molecular scene at
right represents a mixture of
A2 (blue) and B2 (green)
before they react to form AB3.
(a) What is the limiting
reactant?
(b) How many molecules of
product can form?
• The reaction is A2 + 3 B2 → 2 AB3 so the mole ratio
between A2 and B2 is 1:3.
• a) With 3 A2 molecules present, 3 x 3 = 9 B2 molecules
would be required. Since you have only 6 B2 molecules,
B2 is the limiting reagent.
b) 6 B2 molecules x (2 AB3 / 3 B2) = 4 AB3 molecules
3.51 Calculate the maximum numbers of
moles and grams of H2S that can form
when 158 g of aluminum sulfide reacts
with 131 g of water:
Al2S3 + H2O ⟶ Al(OH)3 + H2S
[unbalanced]
What mass of the excess reactant
remains?
3.66 Six different aqueous solutions (with
solvent molecules omitted for clarity) are
represented in the beakers below, and
their total volumes are noted.
A
D
B
E
C
F
(a) Which solution has the highest molarity?
(b) Which solutions have the same molarity?
(c) If you mix solutions A and C, does the
resulting solution have a higher, a lower, or the
same molarity as solution B?
A
D
B
E
C
F
• a) Solution B has the highest molarity as it has the
largest number of particles, 12, in a volume of 50 mL.
• b) Solutions A and F both have 8 particles in a volume of
50 mL and thus the same molarity. Solutions C, D, and
E all have 4 particles in a volume of 50 mL and thus have
the same molarity.
• c) Mixing Solutions A and C results in 12 particles in a
volume of 100 mL. That is a lower molarity than that of
Solution B which has 12 particles in a volume of 50 mL or
24 particles in a volume of 100 mL.
(d) After 50. mL of water is added to solution
D, is its molarity higher, lower, or the same as
that of solution F after 75 mL of water is added
to it?
A
D
B
E
C
F
• d) Adding 50 mL to Solution D would result in 4
particles in a total volume of 100 mL; adding 75
mL to Solution F would result in 4 particles in a
volume of 100 mL. The molarity of each solution
would be the same.
(e) How much solvent must be evaporated from
solution E for it to have the same molarity as
solution A?
A
D
B
E
C
F
• e) Solution A has 8 particles in a volume of 50 mL while
Solution E has the equivalent of 4 particles in a volume of
50 mL. The molarity of Solution E is half that of Solution
A. Therefore half of the volume, 12.5 mL, of Solution E
must be evaporated. When 12.5 mL of solvent is
evaporated from Solution E, the result will be 2 particles in
12.5 mL or 8 particles in 50 mL as in Solution A.
3.69 Calculate each of the following quantities:
(a) Volume in milliliters of 2.26 M potassium
hydroxide that
contains 8.42 g of solute
(b) Number of Cu2 ions in 52 L of 2.3 M
copper(II) chloride
(c) Molarity of 275 mL of solution containing
135 mmol of glucose
3.83 One of the compounds used
to increase the octane rating of
gasoline is toluene (right).
Suppose 20.0 mL of toluene
(d = 0.867 g/mL) is consumed
when a sample of gasoline burns
in air.
(a) How many grams of
oxygen are needed for complete
combustion of the toluene?
(b) How many total moles of
gaseous products form?
(c) How many molecules of water
vapor form?
H
C
NOTE: Complete combustion means CO2
+ H2O as only products