Chem 121 Module 2 Exam
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Portage Learning *
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Course
121
Subject
Chemistry
Date
Jan 9, 2024
Type
Pages
7
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QuesƟon 1
Number your answers, and show your work. Be sure to give answers with the correct units. Calculate the
molecular weight of the following compounds:
1. Ca3N2
2. Li3PO4
3. NH4NO3
Answer:
1. Ca3N2 = 3 (40.08) + 2 (14.01) = 148.26
2. Li3PO4 = 3 (6.941) + (30.97) + 4 (16.00) = 115.793
3. NH4NO3 = 2 (14.01) + 4 (1.008) + 3 (16.00) = 80.052
QuesƟon 2
Number your answers, and show your work. Be sure to give answers with the correct units.
Calculate
the number of moles in 10.0 grams in the following compound:
1. NH4NO3
Calculate the number of grams in 0.0500 mol in the following compound:
2. Li3PO4
Answer:
1. MW of NH4NO3 = 80.052
moles = 10.0 / 80.052 = 0.125 mole
2. MW of Li3PO4 = 115.793
grams = 0.0500 x 115.793 = 5.79 grams
QuesƟon 3
Number your answers, and be sure to show your work. Calculate the % composiƟon of the following
compounds:
1. Na2SO4
2. Ca(NO3)2
Answer:
1. MW of Na2SO4 = 2 (22.99) + (32.07) + 4(16.00) = 142.05
%Na = 45.98 / 142.05 x 100 = 32.37%
%S = 32.07 / 142.05 x 100 = 22.58%
%O = 64.00 / 142.05 x 100 = 45.05%
2. MW of Ca(NO3)2 = (40.08) + 2 (14.01) + 6 (16.00) = 164.1
%Ca = 40.08 / 164.1 x 100 = 24.42%
%N = 28.02 / 164.1 x 100 = 17.07%
%O = 96.00 / 164.1 x 100 = 58.50%
QuesƟon 4
Number your answers, and be sure to show your work. Determine the empirical formula for the
following compound % composiƟons:
1. 52.56% Fe,
2.83% H,
44.91% O
2. 71.98% C, 6.71% H, 21.31% O
Answer:
1) % ComposiƟon of a compound is:
52.56% Fe 52.56% Fe ÷ 55.85 = 0.941 (smallest number of the set)
2.83% H 2.83% H ÷ 1.008 = 2.808
44.91% O 44.91% O ÷ 16.00 = 2.807
0.941 is the smallest of this set of numbers, so it is divided into each of the set of numbers
Fe = 0.941 ÷ 0.941 = 1 Fe
H = 2.808 ÷ 0.941 = 3 H
O = 2.807 ÷ 0.941 = 3 O
FeH3O3
2.) % ComposiƟon of a compound is:
71.98% C 71.98% C ÷ 12.01 = 5.993
6.71% H 6.71% H ÷ 1.008 = 6.657
21.31% O 21.31% O ÷ 16.00 = 1.332 (smallest of the set)
1.332 is the smallest of this set of numbers, so it is divided into each of the set of numbers
C = 5.993 ÷ 1.332 = 4.5
H = 6.657 ÷ 1.332 = 5
O = 1.332 ÷ 1.332 = 1
Since C is 4.5, mulƟply all numbers by 2 to get:
C9H10O2
QuesƟon 5
BALANCE THE FOLLOWING EQUATIONS:
1. SrBr2 + (NH4)2CO3 → SrCO3 + NH4Br
2. FeS2 + O2 → Fe2O3 + SO2
Answer:
1. SrBr2 + (NH4)2CO3 → SrCO3 + 2 NH4Br
2. 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2
QuesƟon 6
Determine which of the following double replacement reacƟons will occur using the solubility rules and
complete the equaƟons that do:
1. Ca+2, 2 Cl-1 + Al+3, 3 NO3-1 →
2. Pb+2, 2 NO3-1 + 2 Na+1, 2 Br-1 →
Your Answer:
1.
Ca^+2, 2 Cl^-1 + Al^+3, 3 NO3^-1 → Will not occur since neither set of ions forms a precipitate
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2.
Pb^+2, 2 NO3^-1 + 2 Na^+1, 2 Br^-1 → PbBr2 ↓ + 2 Na^+1, 2 NO3^-1
QuesƟon 7
Determine which of the following single replacement reacƟons will occur using the acƟvity series and
complete the equaƟons that do:
1. Cl2 + 2 F-1 →
2. Ba + Fe+2 →
Label each of the following reacƟons as either (1) CombinaƟon, (2) DecomposiƟon, (3) CombusƟon, (4)
Double Replacement or (5) Single Replacement and then balance the equaƟon.
3. C6H12O6 + O2 → CO2 + H2O
4. HCl + NaOH → NaCl + H2O
5. Al + Fe(NO3)3 → Fe + Al(NO3)3
6. S + O2 → SO3
7. H2SO4 → SO3 + H2O
Answer:
1. Cl2 + 2 F^-1 → Will not occur since F2 is more acƟve than Cl2
2. Ba + Fe^+2 → Fe + Ba+2 Will occur since Ba is more acƟve than Fe
3. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O CombusƟon
4. HCl + NaOH → NaCl + H2O Double Replacement
5. Al + Fe(NO3)3 → Fe + Al(NO3)3 Single Replacement
6. 2 S + 3 O2 → 2 SO3 CombinaƟon
7. H2SO4 → SO3 + H2O DecomposiƟon
QuesƟon 8
Show the calculaƟon of the oxidaƟon number (charge) of ONLY the atoms which are changing in the
following redox equaƟons and then show how those charges are used to balance the following redox
equaƟons:
1. Mn(NO3)2 + NaBiO3 + HNO3 → HMnO4 + Bi(NO3)3 + NaNO3 + H2O
2. KMnO4 + H2C2O4 + H2SO4 → MnSO4 + K2SO4 + CO2 + H2O
Answer:
1.)
Mn(NO3)2 + NaBiO3 + HNO3 → HMnO4 + Bi(NO3)3 + NaNO3 + H2O
Mn(NO3)2: each NO3 is -1 (total is -2), so Mn is +2
HMnO4: H = +1, each O is -2 (total is -8), so Mn is +7
NaBiO3: Na is metal in group I = +1, each O is -2 (total is -6), so Bi is +5
Bi(NO3)3: each NO3 is -1 (total is -3), so Bi is +3
Since Mn (on leŌ side) is +2 and Mn (on right side) is +7: Mn changes by 5
Since Bi (on leŌ side) is +5 and Bi (on right side) is +3: N changes by 2
MulƟply Mn compounds by 2 and Bi compounds by 5 and aŌer balancing other atoms =
2 Mn(NO3)2 + 5 NaBiO3 + 16 HNO3 → 2 HMnO4 + 5 Bi(NO3)3 + 5 NaNO3 + 7 H2O
2.)
KMnO4 + H2C2O4 + H2SO4 → MnSO4 + K2SO4 + CO2 + H2O
KMnO4: K group I metal = +1, each O is -2 (total = -8), so Mn = +7
MnSO4: each SO4 is -2, so Mn is +2
H2C2O4: each H = +1 (total = +2), each O is -2 (total = -8), so C total is +6, each C is +3
CO2: each O is -2 (total = -4), so C is +4
Since Mn (on leŌ side) is +7 and Mn (on right side) is +2: Mn changes by 5
Since each C (on leŌ side) is +3 and C (on right side) is +4: C changes by 1 x 2 (since H2C2O4) = 2
MulƟply Mn compounds by 2 and H2C2O4 compound by 5 and aŌer balancing other atoms (which
requires placing a 10 in front of CO2) =
2 KMnO4 + 5 H2C2O4 + 3 H2SO4 → 2 MnSO4 + K2SO4 + 10 CO2 + 8 H2O
QuesƟon 9
Complete the following calculaƟons and show your work. Be sure to give answers with the correct units.
1. How many grams of SO2 would be formed from 60 grams of FeS2 in the following reacƟon?
4 FeS2
+
11 O2
→
2 Fe2O3
+
8 SO2
2. How many moles of NH4NO2 would be required to form 25 grams of N2 in the following reacƟon?
NH4NO2
→
N2
+
2 H2O
Answer:
1.
60
X
8
X
64.07
=
64.1 grams of SO2
119.99
4
2.
25
X
1
=
0.892 mol of NH4NO2
28.02
1
QuesƟon 10
Complete the following calculaƟons and show your work. Be sure to give answers with the correct units.
1. Show the calculaƟon of the mass percent of solute in a soluƟon made by dissolving 15.9 grams of
Ca3(PO4)2 in 350 grams of water.
2. Show the calculaƟon of the molality of a soluƟon made by dissolving 15.9 grams of Ca3(PO4)2 in 400
grams of water.
3. Show the calculaƟon of the molarity of a soluƟon made by dissolving 15.9 grams of Ca3(PO4)2 to
make 350 ml of soluƟon.
4. Show the calculaƟon of the mass of Ca3(PO4)2 needed to make 200 ml of a 0.128 M soluƟon.
5. Show the calculaƟon of the volume of 0.238 M soluƟon which can be prepared using 13.4 grams of
Ca3(PO4)2.
Answer:
1. Mass % = (g[solute] / g[solute]+ g[solvent]) x 100% Mass % = (15.9 / 15.9 + 350) x 100 = 4.35%
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2. molality = (g[solute] / MW) / (g[solvent]/ 1000) molality = (15.9 / 310.18) / (400 / 1000) = 0.128 m
3. Molarity = (g[solute] / MW) / (ml[solvent]/ 1000) Molarity = (15.9 / 310.18) / (350 / 1000) = 0.146 M
4. Molarity = (moles) / (ml[solvent]/ 1000) 0.128 = (moles) / (200 / 1000) Moles = 0.128 x 0.200 = 0.0256
Moles = (g[solute] / MW) 0.0256 = (g[solute]/ 310.18) g[solute]= 0.0256 x 310.18 = 7.94 g
5. moles[solute]= g[solute] / MW moles[solute] = 13.4 g / 310.18 = 0.0432 mol Molarity = moles / (mL
/1000) 0.238 = 0.0432 / (mL / 1000) mL / 1000 = 0.0432 / 0.238 = 0.1815 mL = 0.1815 x 1000 = 182 mL
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