CHAPTER 3: ANSWERS TO ASSIGNED PROBLEMS
Hauser- General Chemistry I
revised 9/17/08
3.11 Balance the following equations:
(d) Al4C3 (s) + 12 H2O (l)  4 Al(OH)3 (s) + 3 CH4 (g)
(e) 2 C5H10O2 (l) + 13 O2 (g)  10 CO2 (g) + 10 H2O (g)
do C, H, then half fraction the O2; double everything to remove fraction
(f) 2 Fe(OH)3 (s) + 3 H2SO4 (aq)  Fe2(SO4)3 (aq) + 6 H2O (l)
3.13 Write balanced chemical equations to correspond to each of the following
descriptions:
(b) When solid potassium chlorate is heated, it decomposes to form solid potassium
chloride and oxygen gas.
2 KClO3 (s)  2 KCl (s) + 3 O2 (g)
Δ
(c) Solid zinc metal reacts with sulfuric acid to form hydrogen gas and an aqueous
solution of zinc sulfate.
Zn (s) + H2SO4 (aq)  H2 (g) + ZnSO4 (aq)
3.19 Balance the following equations, and indicate whether they are combination,
decomposition, or combustion reactions:
(a) 2 Al (s) + 3 Cl2 (g)  2 AlCl3 (s)
COMBINATION
(b) C2H4 (g) + 3 O2 (g)  2 CO2 (g) + 2 H2O (g)
COMBUSTION
(c) 6 Li (s) + N2 (g)  2 Li3N (s)
COMBINATION
(d) PbCO3 (s)  PbO (s) + CO2 (g)
DECOMPOSITION
(e) C7H8O2 (l) + 8 O2 (g)  7 CO2 (g) + 4 H2O (g)
COMBUSTION
3.23 Calculate the percentage by mass of oxygen in the following compound:
(a) morphine, C17H19NO3
NOTE: USE 2 DP ATOMIC WTS. AS DISCUSSED
C
H
N
O
17 X 12.01
19 X 1.01
1 X 14.01
3 X 16.00
=
204.17 ÷ 285.37 (X 100%)
=
19.19 ÷ 285.37 (X 100%)
=
14.01 ÷ 285.37 (X 100%)
=
48.00 ÷ 285.37 (X 100%)
sum = 285.37
=
=
=
=
71.5% C
6.7% H
4.9% N
16.8% O answer
sum = 99.9%
PERCENTS OF THIS TYPE ARE TYPICALLY REPORTED TO A TENTH.
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3.33 Calculate the following quantities:
(a) mass, in grams, of 0.105 moles sucrose (C12H22O11)
C 12 X 12.01 =
H 22 X 1.01 =
O 11 X 16.00 =
sum =
144.12
22.22
176.00
342.34 g / mol is formula weight
0.105 mol sucrose ( 342.34 g / mol sucrose) = 35.9457 = 35.9 g (3 SF)
(b) moles of Zn(NO3)2 in 143.50 g of this substance
Zn
N
O
1 X 65.39
=
65.39
2 X 14.01
=
28.02
6 X 16.00
=
96.00
sum = 189.41 g / mol
143.50 g zinc sulfate ( 1 mol zinc sulfate / 189.41 g) = 0.757615 = 0.75762 mol (5 SF)
3.45 Determine the empirical formula of the compound with the following composition
by mass:
Percent to mass
Mass to mole
Divide by small
Multiply till whole
(c) 32.79% Na, 13.02% Al, and 54.19% F
32.79 g Na ( 1 mol Na / 22.99 g)
13.02 g Al ( 1 mol Al / 26.98 g)
54.19 g F ( 1 mol F / 19.00 g)
=
=
=
1.42627 mol Na / 0.48257 = 2.9555 = 3
0.48257 mol Al / 0.48257 = 1.0000 = 1
2.8521 mol F / 0.48257 = 5.9102 = 6
EMPIRICAL FORMULA IS Na3AlF6
3.46 Determine the empirical formula of the compound with the following composition
by mass:
(c) 62.1% C, 5.21% H, 12.1 % N, 20.7% O
62.1 g C ( 1 mol C / 12.01 g )=
5.21 g H ( 1 mol H / 1.01 g ) =
12.1 g N ( 1 mol N / 14.01 g) =
20.7 g O ( 1 mol O / 16.00 g) =
5.1706 / 0.86366
5.1584 / 0.86366
0.86366 / 0.86366
1.2937 / 0.86366
EMPIRICAL FORMULA IS C12H12N2O3
2
=
=
=
=
5.9869 X 2 = 11.973 = 12
5.9727 X 2 = 11.945 = 12
1.0000 X 2 = 2.0000 = 2
1.4979 X 2 = 2.9958 = 3
3.49 Determine the empirical and molecular formula for the following substance:
(a) Styrene, a compound substance used to make Styrofoam cups and insulation,
contains 92.3% C and 7.7% H by mass and has a molar mass of 104 g/ mol.
92.3 g C ( 1 mol C / 12.01 g )=
7.7 g H ( 1 mol H / 1.01 g ) =
7.6852 / 7.6237 = 1.0080 = 1
7.6237 / 7.6237 = 1.000 = 1
EMPIRICAL FORMULA IS "CH"
Since CH has a formula weight of 13.02, divide molar mass by that:
104 g/mol / 13.02 g/mol = 7.99 or 8; 8 will serve as a multiplier
MOLECULAR FORMULA IS C8H8
3.57 Hydrofluoric acid, HF(aq), cannot be stored in glass bottles because compounds
called silicates in the glass are attacked by the HF(aq). Sodium silicate (Na2SiO3), for
example, reacts as follows:
Na2SiO3 (s) + 8 HF (aq)  H2SiF6 (aq) + 2 NaF (aq) + 3 H2O (l)
(a) How many moles of HF are needed to react with 0.300 mol of Na2SiO3?
0.300 mol Na2SiO3 ( 8 mol HF / 1 mol Na2SiO3 ) = 2.40 mol HF (3 SF)
(b) How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?
0.500 mol HF ( 2 mol NaF / 8 mol HF) ( 41.99g NaF / 1 mol NaF) = 5.2487 = 5.25 g
(3 SF)
(c) How many grams of Na2SiO3 can react with 0.800 g of HF?
0.800 g HF ( 1 mol HF / 20.02 g HF) ( 1 mol Na2SiO3 / 8 mol HF) ( 122.07 g / 1 mol
Na2SiO3 ) = 0.6097 = 0.610 g Na2SiO3 (3 SF)
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3.71
Sodium hydroxide reacts with carbon dioxide as follows:
2 NaOH (s) + CO2 (g)  Na2CO3 (s) + H2O (l)
Which reagent is the limiting reactant when 1.85 mol NaOH and 1.00 mol CO2 are
allowed to react?
Choose a product of interest, run both scenarios.
NaOH scenario
1.85 mol NaOH ( 1 mol Na2CO3 / 2 mol NaOH) = 0.925 mol Na2CO3
CO2 scenario
1.00 mol CO2 ( 1 mol Na2CO3 / 1 mol CO2 ) = 1.00 mol Na2CO3
Since the NaOH scenario gives less product, the NaOH is the limiting reactant.
How many moles of Na2CO3 can be produced? 0.925 mol of Na2CO3 can be produced.
The other scenario will not happen.
3.77 When benzene (C6H6) reacts with bromine (Br2), bromobenzene (C6H5Br) is
obtained:
C6H6 + Br2  C6H5Br + HBr
(a) What is the theoretical yield of bromobenzene in this reaction when 30.0 g of
benzene reacts with 65.0 g of bromine?
Do not know the LR. Run both scenarios. Beware of diatomic bromine!
C6H6 scenario
(30.0 g C6H6 ) ( 1 mol C6H6 / 78.12g C6H6) ( 1 mol C6H5Br / 1 mol C6H6)
( 157.01 g C6H5Br / 1 mol C6H5Br) = 60.2956 = 60.3 g C6H5Br (3 SF)
Br2 scenario
65.0 g Br2 ( 1 mol Br2 / 159.80 g Br2 ) ( 1 mol C6H5Br / 1 mol Br2) (157.01 g C6H5Br /
1 mol C6H5Br) = 63.8651 g = 63.9 g C6H5Br (3 SF)
Since the benzene scenario gave less product, this C6H6 is the Limiting Reactant.
The 60.3 g represents the theoretical yield for this reaction.
(b) If the actual yield of bromobenzene was 42.3 g, what was the percentage yield?
% yield = (actual / theoretical) X 100%
(42.3 g actual / 60.3 g theory) X 100% = 70.1%
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