CHEM 1050 Homework
Exam #3 Assignment-Solutions
Alan D. Earhart
7.1
1 mole is equivalent to 6.0221 x 1023 “things”
7.2
6.0221 x 1023
7.4
23
⎛ 4.5 mol Li ⎞ ⎛ 6.0221 × 10 atoms ⎞
= 2.7 × 10 24 atoms Li
a. ⎜
⎟⎠ ⎜
⎟
⎝
1
1 mol
⎝
⎠
23
⎛ 0.0180 mol CO 2 ⎞ ⎛ 6.0221 × 10 molecules ⎞
= 1.08 × 10 22 molecules CO 2
b. ⎜
⎟
⎜
⎟
⎝
⎠⎝
1
1 mol
⎠
⎛ 7.8 × 10 21 atoms Cu ⎞ ⎛
1 mol
⎞
c. ⎜
⎜⎝
⎟ = 0.013 mol Cu
23
⎟
1
⎝
⎠ 6.0221 × 10 atoms ⎠
23
d. ⎛⎜ 3.75 × 10 molecules C2 H 6 ⎞⎟ ⎛⎜
⎝
7.7
1
1 mol
⎞
⎟ = 0.623 mol C2 H 6
23
⎠ ⎝ 6.0221× 10 molecules ⎠
⎛ 1.0 mol quinine ⎞ ⎛ 24 mol H ⎞
= 24 mol hydrogen
a. ⎜
⎟⎠ ⎜
⎝
⎝ 1 mol quinine ⎟⎠
1
⎛ 5.0 mol quinine ⎞ ⎛ 20 mol carbon ⎞
b. ⎜
= 1.0x10 2 mol carbon
⎟⎠ ⎜
⎟
⎝
⎝ 1 mol quinine ⎠
1
⎛ 0.020 mol quinine ⎞ ⎛ 2 mol nitrogen ⎞
= 0.040 mol nitrogen
c. ⎜
⎟⎠ ⎜
⎝
⎝ 1 mol quinine ⎟⎠
1
7.9
a. 188.1 g/mol
d. 342.3 g/mol
b. 159.6 g/mol
e. 58.3 g/mol
c. 329.0 g/mol
7.10
a. 151.9 g/mol
d. 60.0 g/mol
b. 102.0 g/mol
e. 96.0 g/mol
c. 183.1 g/mol
7.14
⎛ 1.50 mol K ⎞ ⎛ 39.1 g K ⎞
= 58.7 g K
a. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molK ⎟⎠
⎛ 2.5 mol C ⎞ ⎛ 12.0 gC ⎞
= 30. g C
b. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molC ⎟⎠
⎛ 0.25 mol P ⎞ ⎛ 31.0 g P ⎞
= 7.8 g P
c. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molP ⎟⎠
⎛ 12.5 mol He ⎞ ⎛ 4.0 g He ⎞
= 50. g He
d. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molHe ⎟⎠
Chapter 7 Solutions
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CHEM 1050 Homework
Exam #3 Assignment-Solutions
Alan D. Earhart
7.15
⎛ 0.500 mol NaCl ⎞ ⎛ 58.5 g NaCl ⎞
= 29.3 g NaCl
a. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molNaCl ⎟⎠
⎛ 1.75 mol Na 2O ⎞ ⎛ 62.0 g Na 2O ⎞
= 109 g Na 2O
b. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molNa 2O ⎟⎠
⎛ 0.225 mol H 2O ⎞ ⎛ 18.0 g H 2O ⎞
= 4.05 g H 2O
c. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molH 2O ⎟⎠
⎛ 4.42 mol CO 2 ⎞ ⎛ 44.0 gCO 2 ⎞
= 194 g CO 2
d. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molCO 2 ⎟⎠
7.16
⎛ 2.0 mol MgCl2 ⎞ ⎛ 95.3 g MgCl2 ⎞
= 190 g MgCl2
a. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molMgCl2 ⎟⎠
⎛ 3.5 mol C6 H 6 ⎞ ⎛ 78.0 gC6 H 6 ⎞
= 270 g C6 H 6
b. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molC6 H 6 ⎟⎠
⎛ 5.00 mol C2 H 6O ⎞ ⎛ 46.0 g C2 H 6O ⎞
= 230 g C2 H 6O
c. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molC2 H 6O ⎟⎠
⎛ 0.488 mol C 3H 6O 3 ⎞ ⎛ 90.0 g C 3H 6O 3 ⎞
= 43.9 g C 3H 6O 3
d. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 molC 3H 6O 3 ⎟⎠
7.24
⎛ 75 g C ⎞ ⎛ 1 molC ⎞
= 6.3 mol C
a. ⎜
⎝ 1 ⎟⎠ ⎜⎝ 12.0 gC ⎟⎠
⎛ 0.25 mol C2 H 6 ⎞ ⎛ 2 mol C ⎞
= 0.50 mol C
b. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 mol C2 H 6 ⎟⎠
⎛ 88 g CO 2 ⎞ ⎛ 1 molCO 2 ⎞ ⎛ 1 mol C ⎞
= 2.0 mol C
c. ⎜
⎟⎠ ⎜
⎝
1
⎝ 44.0 gCO 2 ⎟⎠ ⎜⎝ 1 mol CO 2 ⎟⎠
7.28
a. balanced
7.29
a.
b.
c.
d.
b. balanced
c. unbalanced d. unbalanced
N 2 ( g ) + O 2 ( g ) → 2NO ( g )
2HgO ( s ) → 2Hg ( ℓ ) + O 2 ( g )
4Fe ( s ) + 3O 2 ( g ) → 2Fe 2O 3 ( s )
2Na ( s ) + Cl2 ( g ) → 2NaCl ( s )
Chapter 7 Solutions
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CHEM 1050 Homework
Exam #3 Assignment-Solutions
Alan D. Earhart
7.32
a. Zn ( s ) + 2HNO 3 ( aq ) → Zn ( NO 3 )2 ( aq ) + H 2 ( g )
b. 2Al ( s ) + 3H 2SO 4 ( aq ) → Al2 ( SO 4 )3 ( aq ) + 3H 2 ( g )
c. K 2SO 4 ( aq ) + BaCl2 ( aq ) → 2KCl ( aq ) + BaSO 4 ( s )
d. balanced
7.35
7.49
a. combination
d. double replacement
b. single replacement
e. combustion
c. decomposition
⎛ 2.0 mol H 2 ⎞ ⎛ 1 mol O 2 ⎞
a. ⎜
= 1.0 mol O 2
⎟⎠ ⎜
⎝
1
⎝ 2 mol H 2 ⎟⎠
⎛ 5.0 mol O 2 ⎞ ⎛ 2 mol H 2 ⎞
b. ⎜
= 10. mol H 2
⎟⎠ ⎜
⎝
1
⎝ 1 mol O 2 ⎟⎠
⎛ 2.5 mol O 2 ⎞ ⎛ 2 mol H 2O ⎞
c. ⎜
= 5.0 mol H 2O
⎟⎠ ⎜
⎝
1
⎝ 1 mol O 2 ⎟⎠
7.51
⎛ 0.500 mol SO 2 ⎞ ⎛ 5 mol C ⎞
= 1.25 mol C
a. ⎜
⎟⎠ ⎜
⎝
1
⎝ 2 mol SO 2 ⎟⎠
⎛ 1.2 mol C ⎞ ⎛ 4 mol CO ⎞
b. ⎜
⎟⎠ ⎜⎝
⎟ = 0.96 mol CO
⎝
1
5 mol C ⎠
⎛ 0.50 mol CS2 ⎞ ⎛ 2 mol SO 2 ⎞
= 1.0 mol C
c. ⎜
⎟⎠ ⎜
⎝
1
⎝ 1 mol CS2 ⎟⎠
⎛ 2.5 mol C ⎞ ⎛ 1 mol CS2 ⎞
d. ⎜
⎟⎠ ⎜⎝
⎟ = 0.50 mol CS2
⎝
1
5 mol C ⎠
7.58
75.0 g CaCN 2 ⎞ ⎛ 1 mol CaCN 2 ⎞ ⎛ 3 mol H 2O ⎞ ⎛ 18.0 g H 2O ⎞
a. ⎛⎜
= 50.6 g H 2O
⎟⎠ ⎜
⎝
1
⎝ 80.1 g CaCN 2 ⎟⎠ ⎜⎝ 1 mol CaCN 2 ⎟⎠ ⎜⎝ 1 mol H 2O ⎟⎠
5.24 g CaCN 2 ⎞ ⎛ 1 mol CaCN 2 ⎞ ⎛ 2 mol NH 3 ⎞ ⎛ 17.0 g NH 3 ⎞
b. ⎛⎜
= 2.22 g NH 3
⎟⎠ ⎜
⎝
1
⎝ 80.1 g CaCN 2 ⎟⎠ ⎜⎝ 1 mol CaCN 2 ⎟⎠ ⎜⎝ 1 mol NH 3 ⎟⎠
155 g H 2O ⎞ ⎛ 1 mol H 2O ⎞ ⎛ 1 mol CaCO 3 ⎞ ⎛ 100.1 g CaCO 3 ⎞
c. ⎛⎜
= 287 g CaCO 3
⎟⎠ ⎜
⎝
1
⎝ 18.0 g H 2O ⎟⎠ ⎜⎝ 3 mol H 2O ⎟⎠ ⎜⎝ 1 mol CaCO 3 ⎟⎠
7.62
a. The difference in energies between the products and the reactants
b. exothermic = heat is produced, endothermic=heat is absorbed
c. higher
d. see the second graph on page 240 for the shape and relative energies
Chapter 7 Solutions
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CHEM 1050 Homework
Exam #3 Assignment-Solutions
Alan D. Earhart
7.63
a. exothermic
b. endothermic
c. exothermic
7.109 When oxygen is mentioned in chemical reactions, assume it’s O2
a. In this section, I’ll give you any needed chemical equations. The question in the
book is balanced for one mole of acetylene as follows:
5
C2 H 2 ( g ) + O 2 ( g ) → H 2O ( g ) + 2CO 2 ( g ) + 1300 kJ
2
However, we discussed balancing equations using small whole numbers as
follows:
2C2 H 2 ( g ) + 5O 2 ( g ) → 2H 2O ( g ) + 4CO 2 ( g ) + 2600 kJ
Since this is for 2 moles of C2H2, I needed to double the energy produced.
b. exothermic
⎛ 2.00 mol O 2 ⎞ ⎛ 2 mol H 2O ⎞
= 0.800 mol H 2O
c. ⎜
⎟⎠ ⎜
⎝
1
⎝ 5 mol O 2 ⎟⎠
⎛ 9.80 g C2 H 2 ⎞ ⎛ 1 mol C2 H 2 ⎞ ⎛ 5 mol O 2 ⎞ ⎛ 32.0 g O 2 ⎞
= 30.2 mol O 2
d. ⎜
⎟⎠ ⎜
⎝
1
⎝ 26.0 g C2 H 2 ⎟⎠ ⎜⎝ 2 mol C2 H 2 ⎟⎠ ⎜⎝ 1 mol O 2 ⎟⎠
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