# Solutions

Chapter 1 – Stoichiometry

1.

∴ the empirical formula of the unknown gas is C5O2N3H9

2. (1) Balancing: MnO4 (aq) → Mn2+ (aq)

MnO4 (aq) → Mn2+ (aq) + 4 H2O

MnO4 (aq) + 8 H+ → Mn2+ (aq) + 4 H2O

3 OH + MnO4 (aq) + 8 H+ → Mn2+ (aq) + 4 H2O + 8 OH

8 H2O + MnO4 (aq) → Mn2+ (aq) + 4 H2O + 8 OH

4 H2O + MnO4 (aq) → Mn2+ (aq) + 8 OH

4 H2O + MnO4 (aq) + 5 e → Mn2+ (aq) + 8 OH (x 28)

(2) Balancing C6H12O4 (aq) → 6 HCO3 (aq)

14 H2O + C6H12O4 (aq) → 6 HCO3 (aq)

14 H2O + C6H12O4 (aq) → 6 HCO3 (aq) + 34 H+

34 OH + 14 H2O + C6H12O4 (aq) → 6 HCO3 (aq) + 34 H+ + 34 OH

34 OH + 14 H2O + C6H12O4 (aq) → 6 HCO3 (aq) + 34 H2O

34 OH + C6H12O4 (aq) → 6 HCO3 (aq) + 20 H2O

34 OH + C6H12O4 (aq) → 6 HCO3 (aq) + 20 H2O + 28 e (x 5)

(3) Combine half equations for the final answer: 12 H2O + 28 MnO4 (aq) + 5 C6H12O4 (aq) 28 Mn2+ (aq) + 54 OH + 30 HCO3 (aq)

3.

4.

5. Cl2 (g)

6. +2

7. (1) calculate the number of moles of H3PO4 that would be produced to determine the limiting reagent:

O2 is the limiting reagent

(2) The mass of the excess reagents that did not react:

Excess mass = (205.0 + 225.0 + 240.0) g – 551.2 g = 118.8 g

8. 28.6% by mass means there are 28.6 g C2H6O2 in 100 g of solution. Therefore, the mass of water in the solution is the difference between these values:

 C2H6O2 H2O SOLUTION Mass (g) 28.6 71.4 100 Molar Mass (g/mol) 62.07 Mol 0.461

9.

10.

2 e + Pb2+ (aq) → Pb (s)

2 OH (aq) + CN (aq) → CNO (aq) +H2O (l) + 2 e-

____________________________________________________

CN (aq) + 2 OH (aq) + Pb2+ (aq) → CNO (aq) + H2O (l) + Pb (s)

Reducing agent → CN (aq)Oxidizing Agent → Pb2+ (aq)

11. NO3

12.

Chapter 2 – Gases

1. (1) Balanced chemical equation: C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l)

(2) Find the mass of CO2 (g):

• O2 is the limiting reagent ∴ mass of CO2 = (1.850 mol)(44.01g/mol) = 81.4 g

(3)

(4)

2. (1)

∴ the empirical formula of the unknown gas is C8O3N5H21

(2)

∴ the molecular formula of the unknown gas is C24O9N15H63

3.

4. (a)

? mass O2 = (2.770 mol)(32.00 g/mol) = 88.7 g

(b)

5. (1)

Find the initial temperature:

6. 0.020 mol/hr

7. False

8. (a)

(b)

9.

10.

Chapter 3 – Thermochemistry

1. (1) calculate the value of q for the combustion of 1.22 g of C6H10O(l) at a constant volume (which means that q = ∆U)

qreaction = ΔU = – qwater – qcal

qreaction = – mH2O∙c∙∆TH2O-(Ccal∙∆Tcal)

qreaction = -(2725 g)(4.188 J/g∙K)(2.75 K) – (3500 J/K)(2.75 K)

qreaction = – 40988 J

qreaction = – 40.988 kJ

For one mole; there is 98.14 g

x = ΔU = – 3297 kJ

(2) now, we will find Q, W, ∆H and ∆U at constant pressure. But ∆U is always -3297 kJ

2. (a)

(b)

3. The heat that leaves the iron enters the water and the container:

4. (1) Use Hess’ Law

C (graphite) + O2 (g) → CO2 (g)                                      ΔH = – 393.5 kJ

2 H2 (g) + O2 (g) → 2 H2O (l)                                           ΔH = 2(- 285.8 kJ) = – 571.6 kJ

CO2 (g) + 2 H2O (l) → CH3OH (l) + 32 O2 (g)                 ΔH = -(- 726.4 kJ) = +726.4 kJ

CH3OH (l) → CH3OH (g)                                                  ΔH= + 37.4 kJ

C (graphite) + 2 H2 (g) + 12 O2 (g) → CH3OH (g)                   ΔH = -201.3 kJ

(2)

5.

Technically, the sign here should be negative since heat is leaving, but since this is already acknowledged in the question, it’s not needed.

6. q = mc∆T = – 2.75 × 104 kJ

7.  (a)

(b)

8.

9.

Chapter 4 – Chemical Equilibrium

(b)

 2 A (g) 2 B (g)                     + C (g) I 4.00 atm 2.00 atm C -2x + 2x +x E 4.00 – 2x 2.00 + 2x x

2.

 A (aq)                   + B (aq)         ⇌ 2 C (aq) I 0.322 M 0.244 M 0.455 M C -x -x +2x E 0.322 – x 0.244 – x 0.455 + 2x
3.854x2 + 1.903x + 0.1955 = 0  →  x = – 0.146 and – 0.348

(note that x is found using the quadratic formula and x = -0.348 is impossible)

[C] = 0.455 + 2x = 0455 + (2)(- 0.146) = 0.164 M

3. Note that the molar mass of AgNO3 is 169.87 g/mol

 Ag+ (aq)                     + 2 CN– (aq)                    ⇌ Ag(CN)2– (aq) I 0.03903 0.800 C -0.03903 -20.03903 +0.03903 E 0.72194 0.03903

Note: a small amount of Ag(CN)2will react to regenerate Ag+ … the concentrations of  CN and Ag(CN)2are not affected.

4. (a)

 A (g) 2 B (g) 2 C (g) I 0.500 0.500 0 C -0.245 -0.245 +0.489 E 0.255 0.011 0.489

(b)

5. Ptotal = 20.0 bar = PCH4 + PCO2. Since nCH4 = nCO2 (equimolar), that means each gas has an initial partial pressure of 10.0 bar.

 CH4 (g) CO2 (g) 2 CO (g) 2 H2 (g) I 10.0 10.0 0 0 C -x -x +2x +2x E 0.246 0.2 19.6 19.6

6. 0.83

Divide Equation 1’s equilibrium constant by 2 and multiple Equation 2’s equilibrium constant by -1 (as you have to flip it); Add modified equilibrium constants to get 0.83.

Chapter 5 -Acid/Base Equilibria

1. NH4+ (aq)          ⇌          NH3 (aq)    +    H+ (aq)

E:  0.333 – x 0.333                x                      x

pH = -log(1.36 x 10-5) = 4.87

2. (a)

(b) HA (aq)          ⇌    H+ (aq)     +       A(aq)

E:   0.1971 – x                  x                      x

(c) A(aq)          +        H2O (l)      ⇌      HA (aq)     +     OH(aq)

E:   2.000 – x 2.000                                      x                         x

pH = 8.19

3. (a)

 2 A (aq)                ⇌ B (aq)              + C (aq) I 0.444 0.555 0.666 C -2x +x +x E 0.444 – 2x 0.555 + x 0.666 + x

21.2x2 – 11.0778x + 0.7245 = 0  →    x = 0.4459 and 0.0765 (Note: use quadratic formula to find values for x and x = 0.4459 is impossible so we use 0.0765)

[A] = 0.44 – 2(0.0765) = 0.291 M

(b)

∴ pOH = 2.29 and pH = 11.71

4.

pOH = 14 – pH – 3.45

[OH-] = 10-3.45 = 3.55 x 10-4 M

Let B represent the base, N(CH3)3,

 B H2O OH– HB+ I [B] – 0 0 C -x – +x +x E [B] – x – x X

[B] = 0.0200 M

Check:

5.  (a) Initial pH

 HA H2O H3O+ A– I 0.090 – 0 0 C -x – +x +x E 0.090 – x – x X

Check:

(b) At half equivalence point: pH = pKa = -log(6.2 x 10-10) = 9.21 (after 40 mL added)

(c) At the equivalence point

mol HA = mol OH- added = 0.0072 mol (this corresponds to adding 80.0 mL of base)

new [A-] = 0.0072 mol/(0.080 L + 0.080 L) = 0.0450 M

OCl is a conjugate base of a weak acid, so it hydrolyzes:

 A– H2O HA OH– I 0.045 – 0 0 C -x – +x +x E 0.045 – x – x x

x = 8.5 x 10-4 M = [OH-]

pOH = -log(8.5 x 10-4) = 3.07

pH = 14 – 3.07 = 10.93

(d)

Chapter 6 – Ionic Equilibria in Aqueous Systems

1. (a) The molar mass of Mg(PO4)2 is 262.86 g/mol
 Mg(PO4)2 (s)  ⇌ 3 Mg+ (aq)       + 2 PO43- (aq) I C + 3x + 2x E 3x 2x

(b)

 Mg(PO4)2 (s) ⇌ 3 Mg+ (aq)       + 2 PO43- (aq) I 0.30 C + 3x + 2x E 0.30 + 3x 0.30 2x

2. (a)

 NH3 (aq)           + H+ (aq)              ⇌ NH4+ (aq) I 0.7103 0.1701 C – 0.1701 – 0.1701 + 0.1701 E 0.5402 0.1701

(b) 1.00 g of NaOH (0.0250 mol) consumes 0.0250 mol of NH4+ (aq) and produces 0.0250 mol of NH3 (aq)

(c) 1.00 g of HCl (0.0274 mol) consumes 0.0274 mol of NH3 (aq)and produces 0.0274 mol of NH4+ (aq)

3. The Ba(OH)2 has 2 groups of OH, so CAVA = 2 CBVB

• At the equivalence point, the volume is (70.4 mL + 28.0 mL) = 98.4 mL
• All of the acetic acid is converted to CH3COO; a weak base:

pH = 9.02

4.

pH = 14 – pOH   pH=8.88

5. (a)

(b)

To lower pH to 9.30, we need to add acid. But how much?

Amount: 0.41 g

6. (a) PbI2

(b)

 Pb2+ I- PbI2 (s) B 8.494 x 10-2 mol 0 A 3.26 x 10-5 mol M -0.5 x 3.26 x 10-5 mol -3.26 x 10-5 mol +0.5 x 3.26 x 10-5 mol A 8.494 x 10-2 mol 0 1.63 x 10-5 mol

1.63 x 10-5 mol of PbI2 precipitates and total volume is 20.0 mL + 30.0 mL = 51.0 mL

? [Pb2+] = 8.494 x 10-2 mol/0.0510 L = 1.665 M

? [NO3] = 0.1699 mol/0.0510 L = 3.33 M

? [Na+] = 3.26 x 10-5 mol/0.0510 L = 6.39 x 10-4 M

? [I-] = 0 mol/0.0510 L = 0 M

7. Circled → Ca(CN)2 and LiF

Underlined → KCl and CuNO3