Baixe Resoluções de alguns capítulos do livro de balanço de massa e conversão de unidade e outras Resumos em PDF para Calor e Transferência de Massa, somente na Docsity!

2.1. (a) N/mm or nm (nanometer) (b) °C/M/s (c) 100 kPa (d) 273.15 K (e) 1.50m, 45 kg (f) 250°C (g) J/s (h) 250 N

2.2.1 (^) a. Basis: 1 mi 3 1 mi^3 5280 ft 1 mi

(^3 12) in 1 ft

(^3 2). 54 cm 1 in

(^3 1) m 100 cm

3

= 4.17 × 10 9 m 3

b. Basis: 1 ft^3 /s 1 ft 3 1 s

60 s 1 min

7.48 gal 1 ft 3

= 449 gal / min

2.2.

a.

0.04 g

(min ) (m 3 )

60 min 1 hr

(12 in )^3

1 ft 3

1 lb (^) m 454g

lb (^) m

(hr ) (ft 3 )

b.

2L

s

3600 s 1 hr

24 hr 1 day

1 ft 3 28.32L

= 6.1× 10 3

ft 3 day

c. 6 (in.)(cm 2 ) (yr)(s)(lb (^) m )(ft 2 )

1 ft 2 (12 in) 2

(1 in) 2 (2.54 cm) 2

1 yr 365 days

1 day 24 hr

1 hr 3600s

2.2 lb (^) m 1 kg

1 ft 12 in

0.3048 m 1 ft

= 1.14^ ×^10

• 1 1 m

(kg ) (s 2 )

2.2.3 a. Basis: 60.0 mile/hr 60.0 mile 5280 ft 1 hr (^) = 88ft hr 1 mile 3600 sec sec

b. Basis: 50.0 lb (^) m /(in) 2 2 2 m 4 2 2 2 2

50.0/lb 454 g 1 kg 1(in) (100 cm) (^) = 3.52 10 kg (in) 1 lb 1000 g (2.54 cm) (1 m) m

×

c. Basis: 6.20 cm/(hr) 2 9 2 2 2 2

6.20 cm 1 m 10 nm 1(hr) nm = 4. (hr) 100 cm 1 m (3600 sec) sec

2.2.4 (^) 20 hp 0.7457 kW 1 hp = 14.91 kW

No , not enough power even at 100% efficiency; 68 kW = 91.2 hp.

2.2. 1 hr 525 mile

2200 gal 1 hr

1000 mile = 4190.5 gal

1 hr 475 mile

2000 gal 1 hr

1000 mile

4210 gal (20 gal)

None: 20 gal more are needed.

2.2.6 Let t A be the time for A to paint one house; t B for B A does a house in 5 hours, or 1 house/5 hr. B does one house in 3 hours, or 1 house/3 hr.

1 house (^) A hr (^) + 1 house (^) Bhr= 1 house 5 hr 3 hr

t t

Also t A = t B so that (^) A A A

• = 1 or = 15 15 15

t t t

A B

= hr = = 1.875 hr or 112.5 min 8

t t

Solutions Chapter 2

2.2.7 (a) mass, because masses are balanced (b) weight, because the force exerted on the mass pushes a spring

2.2.8 (^) 20.0g

(m)(s)

1 lb (^) m 453.6 g

0.3048m 1 ft

3600s 1 hr

1 (lb (^) f )(s 2 ) 32.174(lb (^) m )(ft)

1(hr) 2 (3600) 2 s 2

= 1.16 × 10 −^7 (lb (^) f )(hr) ft 2

2.2.9 (^) 1.0 Btu

(hr)(ft 2 )

o (^) F ft

24 hrs 1 day

1 ft 2 (12 in) 2

1 in 2 (2.54 cm 2 )

(100 cm) 2 1 m 2

1.8 o^ F 1 o^ C

2.54 cm in

252 cal 1 Btu

12 in 1 ft

4.184 J

1 cal

1 kJ 1000 J

= 1.49 × 10 4

kJ (day)(m 2 )(ϒC / cm)

2.2.10 Basis: 1 lb H 2 O

a.

(^3 ) 2 m f f m

KE= 1 mv = 1 1 lb^ 3 ft^ (s )(lb ) = 0.14(ft)(lb ) 2 2 s 32.174(ft)(lb )

b. Let A = area of the pipe and v = water velocity. The flow rate is

q = Av = πD 2 4

⎠⎟^

(v)

π 4

(2 in) 2 (1 ft) 2 (12 in) 2

3 ft S

60 s 1 min

7.48 gal 1 ft 3

= 29.37 gal/min

2.2. PE = 75 gal min

8.33 lb (^) m gal

32.2 ft sec 2

ft

60 s hr

s 2 -lb (^) f 32.2 lb (^) m -ft

Btu 778 ft-lb (^) f = 4818 Btu/hr

2 hp 2545 Btu Pump Work 5090 Btu/hr hp-hr

Rate of energy input for heating = PW - PE =5090 - 4818 = 272 Btu/hr

Solutions Chapter 2

2.2.12 The object has a mass of 21.3 kg (within a precision of ± .1 kg). The weight is the force used to support the mass.

2.2.13 In American Engineering System Power = FV

= 800 lbf^ 300 ft^ = 2.4 105 (lb )(ft)f or 7.27 hp 1 min min

×

In SI

Power = 4000 N 1.5 m 1 (watt)(s) 1 s 1(N)(m)

= 6000 watts

2.2. KE = 1 m v^2 2 =

1 2

2300 kg 1 lb (^) m

1. 454 kg

2. 0 ft s

32. 2 lb (^) m lb (^) f

ft sec 2

1 Btu

778. 2 (ft )( lb (^) f)

= 10.11 Btu

2.2.15 Basis: 10 tons at 6 ft/s

KE =

(^2 ) (^2) m f f m

1 1 20,000 lb 6 ft 1(s )(lb ) m v = = 11,200(ft)(lb ) 2 2 s 32.2(ft)(lb )

2.2.16 (^) Basis: 1 mRNA 1 nRNA 3 x ribonucleotides nRNA

1 active ribosome 264 ribonucleotides

1200 amino acids min-active ribosome

protein x amino acids =

13.6 protein molecules formed per min per nRNA

Denominator = (^) ΔHccat

= energy mol substrate C

(2)

η = (1) energy (2) energy

There is no missing conversion factor What the author claimed about the units is correct.

2.3.9 (^) For dimensional consistency: B – absolute temperature in either ºR or K C – absolute temperature in either ºR or K A – dimensionless In most cases the argument of a logarithm function should be dimensionless, but in this case it will not be. Therefore, the numerical values of A, B, and C will depend upon the units used for temperature and the units used for p *.

2.3.10 The equation is

Δp = 4fρ ⎡⎣(v 2 / 2g) (L/D)⎤⎦

The units on the right hand side (with f dimensionless) in SI are

ρ kg m 3

m 2 s 2 (kg)(m) s 2

m m

m 2

hence the equation is not dimensionally consistent because Δ p has the units of N/m 2. If g is replaced with g, the units would be correct.

2.4.1 Two because any numbers added to the right hand side of the decimal point in 10 are irrelevant.

2.4.2 The sum is 1287.1430. Because 1234 has only 4 significant figures to the left of an implied decimal point, the answer should be 1287 (no decimal point).

2.4.3 The number of significant figures to the right of the decimal point is 1 (from 210.0m), hence the sum of 215.110 m should be rounded to 215.1 m.

2.4.4 A calculator gives 569.8269 000, but you should truncate to 4 significant figures, or 569.8 cm 2.

2.4.5 Two significant figures (based on 6.3). Use 4.8^ ×^10 .

2.4.6 Step 1: The product 1.3824 is rounded off to 1. Step 2: Calculate errors.

For absolute error, the product 1.4 means 1.4 + 0.

Thus 0.05^ 100% 3.6%

× = error

Similarly 3.84 has 0.005^ 100% 0.13%

× = error

and 0.36 has

100% 1.4%

× = error

Total 2.7% error

2.6. a) =

b) =

c) = 1.26 × 10 −^3 lb mol N 2

(d) (^2 6 2 62 ) 2 6

3 lb C H O 1 lb mol C H O 454 g mol = 29.56 g mol C H O (46.07) lb C H O 1 lb mol

4 g mol mg Cl 2 (95.23 )g MgCl 2

gmol MgCl (^2)

380.9 g

2 lb mol C 3 H 8 ( 44.09)lb C 3 H 8

lb mol C 3 H (^8)

454g C 3 H (^8) 1 lb C 3 H (^8)

4 × 10

4 g C 3 H (^8)

16 g N 2 gmol N (^2)

(28.02 )g N 2

1 lb mol N (^2) 454 gmol N (^2)

Solutions Chapter 2

2.6. (a) 16.1 lb mol HCl^ 36.5 lb HCl^ = 588 lb HCl 1 lb mol HCl

(b) 19.4 lb mol KCl 74.55 lb KCl = 1466 lb KCl 1 lb mol KCl

(c)

3 3 3 3

11.9 g mol NaNO 85 g NaNO 2.20 10 lb = 2.23 lb NaNO 1 g mol NaNO 1 g

×

(d)

2 2 2 2

164 g mol SiO 60.1 g SiO 2.20 10 lb = 21.7 lb SiO 1 g mol SiO 1 g

×

2.6.3 Basis: 100 g of the compound

comp. g MW g mol Ratio of Atoms C 42.11 12 3.51 1. O 51.46 16 3.22 1 H 6.43 1.008 6.38 2

Multiply by 11 to convert the ratios into integers

The formula becomes C 12 O 11 H (^22) Checking MW: 12(12) + 11(16) + 22(1.008) = 342 (close enough)

Solutions Chapter 2

2.6.4 (^) Vitamin A, C 20 O H 30 , Mol Wt.: 286 Vitamin C, C 6 H 8 O 6 , mol. wt: 176

Vitamin A:

a. Vitamin A = 2.00 g mol 286 g 1 lb = 1.26 lb 1 g mol 454 g

16 g 1 lb 0.0352 lb 454 g

Vitamin C = 2.00 g mol 176 g 1 lb = 0.775 lb g mol 454 g

16 g 1 lb 0.0352 lb 454 g

b. Vitamin A = 1.00 lb mol 286 lb 454 g = 130,000 g 1 lb mol 1 lb

Vitamin C = 1.00 lb mol^ 176 lb^ 454 g = 79,900 g 1 lb mol 1 lb

For both 12 lb 454 g 5450 g 1 lb

2.6.5 (^) 1 kg 5.2 m 3

1160 m 3 kg mol = 223.1 kg/kg mol

2.6.6 (^) Mass fraction to mole fraction:

x 1 =

ω 1 MW 1 ω 1 MW 1

(1− ω 1 ) MW 2

Mole fraction to mass fraction

ω 1 = x 1 MW 1 (x 1 )(MW 1 ) + (1− x 1 )(MW 2 )

2.6.7 (^) Basis: 100 kg mol gas

Comp. Mol % = mol MW kg CH 4 30 16 480 H 2 10 2 20 N 2 60 28 1680 Total 100 2180 21.8 kg kg mol

2.6.8 (^) Basis: 100 g mol gas

Comp. mol = % MW g CO 2 19.3 44 849. N 2 72.1 28 2018. O 2 6.5 32 208. H 2 O 2.1 18 37. 100.0 3113.

Avg. mol. wt = 31.

2.6.9 (^) Basis: 100 lb mol

Comp. % = mol MW lb CO 2 16 44 2640 CO 10 28 280 N 2 30 28 840 3760

Avg. MW = 37.6 lb/lb mol

2.7.1 (a) A gas requires a convenient basis of 1 or 100 g moles or kg moles (if use SI units). (b) A gas requires a convenient basis of 1 or 100 lb moles (if use AE units). (c) Use 1 or 100 kg of coal, or 1 or 100 lb of coal because the coal is a solid and mass is a convenient basis.

(d) Use 1 or 100 moles (SI or AE) as a convenient basis as you have a gas. (e) Same answer as (e).

2–

2.7.2 Since the mixture is a gas, use 1 or 100 moles (SI or AE) as the basis.

2.7.3 Pick one day as a basis that is equivalent to what is given—two numbers: (a) 134.2 lb C1 (b) 10.7 × 10 6 gal water.

2.8.1 Basis: 1000 lb oil

= 129.5 gal

2.8.2 Basis: 10,010 lb

2.8.3 Basis: 1 g mol each compound g mol mw g ρ (g/cm 3 ) V (cm 3 )

Pb 1 207.21 207.21 11.33 18.

Zn 1 65.38 65.38 7.14 9. C 1 12.01 12.01 2.26 5.

3 ˆ mass (g) V = density (g/cm )

Set 2 is the correct one

2.9.1 (^) Component g mol mol fraction MW g mass fraction Na 1 0.20 23 23 0. C1 1 0.20 35.45 35.45 0. O 3 0.60 16 48 0. 5 106.45 1.

0.926 lb oil ft 3 1.00 lb H^2 O ft 3

62.4 lb H^2 O ft (^3) = 57.78 lb oil ft 3 oil

1000 lb oil 1 ft 3 oil 57.78 lb oil

7.48 gal 1 ft 3

10,010 lb 1 gal 8.80 lb

0.134 ft^3 gal

152 ft 3

Solutions Chapter 2

2.9.2 (^) Basis: 1 gal of solution.

Mass of solution:

M ass fraction KOH = =

Mass fraction H2O = 1 – 0.09 =

2.9.3 (^) Basis: 30 lb gas

Comp. lb MW lb mol mol fr. CO 2 20 44 0.455 0. N 2 10 28 0.357 0. 30 0.812 1.

2.9.4 (^) a) 1000

b) No

c) Yes, because for solids and liquids the ratio in ppb is mass, whereas for gases the ratio is in moles.

1.0824 lb soln ft 3 H 2 O

lb H 2 O ft 3 H 2 O

lb H 2 O ft 3 H 2 O 1 ft 3

7. 481 gal

1 gal 9.03 lb soln

0.813 lb 9.03 lb

1.00 Total

Solutions Chapter 2

2.9.5 (^) On a paper free basis the total ppm are:

Brand A: 6060 ppm Brand B: 405 ppm

The respective mass fractions are:

The other entries are similar Fe Cu Pb

Brand A:

Brand B:

Ends

= 54.4 m^3

Vs = 5 m 30.2 m 200 mm 1 m 1000 mm

2 sides = 60.4 m 3

Ve = 5 m 27.2 m 200 mm 1 m 1000 mm

2 ends

2.9.15 No. 1 molecule in 10 23 or more is not 13-20 ppb

2.9.16 On a mol basis, the carbon dioxide concentration in air is about 350 parts per million (ppm), while that of oxygen is about 209,500 ppm. If the atmospheric concentration of carbon dioxide is increasing at about 1% per year (i.e., from 350 ppm this year to 353.5 ppm next year), and not to 1%, the 3.5-ppm change in dioxide concentration causes the oxygen concentration to fall from 209,500 to about 209,497 ppm, which is less than a 0.002% decrease. So, there is no need to worry about an oxygen deficit at present.

2.10.1 TK = –10 + 273 = 263K T°F = –10 (1.8) + 32 = 14°F T°R = 14 + 460 = 474°R

2.10.2 Yes, if the temperature scale is a linear relative one (°C, °F), or a logarithmic scale (ln 1° is zero). No, if the scale is absolute, but read J. Wisniak, J. Chem. Educ ., 77 , 518-522 (2000) for a different conclusion.

2.10.3 C

p = 8.41 +^ 2.4346^ ×^10

-5 T

K ⎛2.4346× 10 -5 (^) (gmol)(K)J 2 ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟(TK )

Substitute 1.8 TK = T°R

Cp =

= 8.41 +1.353 × 10 −^5 ToR

8.41 + 2.4346 × 10 œ^

J

(gmol ) (K )^2

TϒR

2.10. a) + 32 =

b) + 32 + 460°R =

c) − 25 o^ F – 32 o^ F

1.0 o^ C 1.8 o^ F

+ 273K = 241.3K

d) =

2.10.5 First multiply the RHS of the equation so that

Btu 1054.8J 1 lb mol 1.8°F 1°C (lbmol)(°F) 1 Btu 454 gmol 1.0°C 1K

= 4.

and substitute T°F = 1.8T°C + 32

C (^) p = [8.488 + 0.5757 × 10 −^2 (1.8T°C + 32) − 0.2159 × 10 −^5 (1.8T°C + 32) 2 + 0.3059 × 10 −^9 (1.8T°C + 32) 3 ]4.

J

(gmol)(°K)

Simplifying,

Cp = 36.05 + 0.0447T – 0.2874 10 –4 T^2 + 0.7424 10 –8 T^3

2.10.6 The instrument does not contain mercury, but has to contain a fluid that responds at –76°C and can be calibrated to measure temperature.

10 ° C 1.8° F

1.0° C 50 ϒF

10 ϒC 1.8ϒF

1.0ϒC 510 ϒR

150K 1.8ϒ R

1.0K 270 ϒR

J

(gmol )( ϒ K)

× ×

Solutions Chapter 2

2.11.1 (^) Basis: 15cm^3 water

a. (^3) 1000 kg 10 m 10 m 0.15 m m = ρV= = 15,000 kg m

b. (^2) 2 2

F (^) = m g (^) = 15,000 kg 9.80 m 1 N 1 Pa= 1470 Pa A A s 10 m 10 m (kg)(m)^ N 1 s m = 1.47 kPa

1.470 kPa 1 atm 14.7 psi = 0.21 psi 101.3 kPa atm

or 2 2

15 cm H O 1 in. 1 ft 14.696 psi (^) = 0. 2.54 cm 12 in. 33.91 ft H O

2.11.2 (^) ρ concrete =^ 2080 kg/m^ 3 ρwater = 1000 kg/m 3

Volume of concrete:

Sides

Ends

= 54.4 m^3

Floor

Vs = 5 m 30.2 m 200 mm 1 m 1000 mm

2 sides = 60.4 m 3

Ve = 5 m 27.2 m 200 mm 1 m 1000 mm

2 ends

Vf = 27.4 m 30.4 m 200 mm 1 m 1000 mm = 166.592 m 3

Solutions Chapter 2

Total volume = 281.392 m^3

Mass of concrete =

3 3

281.392 m 2080 kg = 585, m

Volume of displaced water required to float:

V = LWh

h = 0.703 m

2.11.3 The pressure is a gauge pressure.

Basis: 50.0 psig

a.

(difference)

b. No. Insufficient height.

Alternate solutions can be applying ∆p =

2.11.4 2 f 2 f^ m

lb in. = lb lb in the AE system in.

so the procedure is ok, although the unit conversion is ignored.

586,543.36 kg 1 m 3 H 2 O 1000 kg H 2 O = 586.543 m 3

→ h =

V

LW

586.54 m 3 27.4m 30.4 m

50.0 psig 33.91 ft H 2 O 14.7 psia = 115 ft

ρΔhg

ft

60°F

2.11.5 Basis: Dept = 1000 m

p = 1 atm +

Alternative solution:

2 2 4 2 2

1000 m sea H O 1.024 g H O 3.28 ft 101.3 kPa 101.3 kPa + = 1.003 10 1.00 g sea H O 1 m 33.91 ft H O ×

4

• 0.01 104 = 1.013 10

× ×

2.11.6 (^) (a) p (^) ABS = p (^) Gauge + patm 2 2 f 2 2 2 2 f

22.4 lb 144 in. 28.6 in. Hg 14.7 psia 144 in.

• = 5250 lb /ft 1 in. 1 ft 1 29.92 in. Hg 1 ft

⎜ ⎟⎜^ ⎟

(b) 22.4 psig 29.92 in. Hg 14.7 psia

⎠⎟^

• 28.6 in. Hg = 74.2 in. Hg

(c) 22.4 psig 1

1.013 × 10 5 N/m 2 14.7 psia

⎠⎟^

• 28.6 in. Hg 1

1.013 × 10 5 N/m 2 27.92 in. Hg

⎠⎟^

= 2.51× 10 5 N/m 2

2 (d) 22.4 psig 33.91 ft H O^2 + 28.6 in. Hg 33.91 ft H O^2 = 84.1 ft H O 1 14.7 psia 1 29.92 in. Hg

2.11.7 Neither John is necessarily right. The pressure at the top of Pikes Peak is continually changing.

p = p (^) o + ρgh

1000 m 1.024 g cm 3

9.8 m s 2

1 kg 103 g

100 cm 1 m

3 1 N

(kg ) (m )

s 2

1 kPa 1N

(kg ) (m )

s 2

1 atm 1.013 × 10 5 N / m 2

100.1 atm 101.3 kPa 1 atm

= 1.014^ ×^10

(^4) kPa

2.11.

a.

b.

Δp = ρgh

= 1 g cm 3

9.8 m s 2

13.1 m 1 kg 1000 g

(100 cm )^3 m^3

Pa 1 kg m s^2

1 kPa 1000 Pa

= 128. 4 kPa F A

mg A (^) Bottom

ρVg A (^) Bottom

ρ( S.A. ) ( )tg

A (^) Bottom

A (^) Bottom = π 4

D 2 =

π 4

(30.5 m )^2

= 730.62 m 2

ATop = 730.62 m^2

A side = πDh = π (30.5 m )( 13.1m ) = 1255.2 m 2

S.A.= A (^) Top + A (^) Bottom + A (^) Side

= (730.62 + 730.62 +1255.22) m

2 = 2716.46 m 2

ρ (S.A. )( ) tg

A

7.86 g cm 3

(100 cm 3 )

m 3

2,716.46 m 2 9.35 × 10 œ^3 m 730.6 m 2

9.8 m s 2

1 kg 1000 g

1 Pa 1 kg / ms 2

1 kPa 1000 Pa = (^) 2.68 kPa

Solutions Chapter 2

2.11.

p -p A B = ρ 1 gh + 1 ρ 2 gΔh ρ− 1 gh 1 ρ− 1 gΔ h

= ρ 2 gΔh 1 −ρ 1 gΔh = gΔh(ρ 2 −ρ 1 )

=18.4 mm Hg

A 1 1 2 D B 1 1 2 D

p + ρ gh + ρ gΔh = P p + ρ gh + ρ gΔh = P

The pressure at A is higher than the pressure at B.

Alternate solution:

Δp = (0.78 in.) (13.546^ −^ 0.91)

760 mm Hg 29.92 in. Hg

= 18.5 mm Hg

= 9.8 ms 2 0.78 in^ 3.28 ft1 m13.546 œ0.91gcm 2 1000 g1 kg

1 ft 12 in

100 cm 1 m

3

1 N

m 2 1 kg

(m ) ( )s^2

1 Pa 1 N m 2

760 mm H (^2) 101.3 × 10 3 Pa

Solutions Chapter 3

3.1. Boundary: Around both pumps and include the soil at the end of the pipes.

It’s an open system.

It’s at steady-state except at startup (note system boundary limits fluid), or if fluid enters, unsteady state.

3.1.2 Either open (flow) or closed (batch) is acceptable if explanation is given

(1) flow – material comes in and out continuously over a suitably long period of time

(2) batch – material is injected into the system, and then in a short period of time, a reaction occurs with the system valves closed.

3.1. (a) open if you have to replace water, and water evaporates; otherwise closed

(b) open

3.1. If the overall flows in and out over a time period for several batches are considered, and the local batches ignore, the process can be treated as continuous.

3.1. a) closed

b)open

c)open

d)closed

3.1. The density of the crystalline silicon in the cylinder is 2.4 g/cm^3.

Basis: 62 kg silicon

The system is the melt, and there is no generation or consumption. Let Δmt be the accumulation.

Accumulation Input Output Δmt = 0 - 0.5(62 kg)

Δm (^) t = −31 kg Let t be the time in minutes to remove one-half of the silicon 2 3

2.4 g π(17.5cm) 0.3 cm min 1 = (62,000 g) cm 4 min 2

t

t = 179min

3.1. Basis: 1 hr

Overall material balance (kg): 13,500 + 26,300? 39,

Yes. The balance is satisfied

NaC1 balance: 0.25 (13,500) + 0.05 (26,300)? 0.118 (39,800) 4690 ≅ 4696 The balance is closely satisfied but not exactly. The closure is good for industrial practice.

3.1. Basis: 1 hour

The overall material balance is

In (lb) Out (lb) 106,000? 74,000 + 34,000 = 108,

The error in the overall material balance is 2000 lb/h or 1.9%; therefore, the overall material balance is within the expected error for industrial flow sensors.

Propylene balance: 0.7×106,000? (0.997) (74,000) + (0.1) (34,000) 74,200? 73,778 + 3,400 = 77,178 (3.8% error) Propane balance: 0.4×106,000? (0.003) (74,000) + (0.9) (34,000) 31,800? 222 + 30,600 = 30,622 (3.8% error)

Note that the relative error for the component balances are twice as large as the relative error for the overall material balance, indicating that there is additional error in the composition measurements used for the component balances.

3.1. Basis: 100 kg wet sludge

The system is the thickener (an open system). No accumulation, generation, or consumption occur. The total mass balance is

In Out

100 kg 70 kg + kg of water

Consequently, the water amounts to (^) 30 kg.

3.2.1 (a) water and air. (b) Insulation, air and what is in the atmosphere. (c) Yes (cold water in, hot water out) (d) Yes

Solutions Chapter 3

3.2.

Four balances are possible, 3 components plus 1 total.

0.10F 1 + 0.50F 2 + 0.20F 3 = 0.35P 0.20F 1 + 0F 2 + 0.30F 3 = 0.10P 0.70F 1 + 0.50F 2 + 0.50F 3 = 0.55P

Total balance F 1 + F 2 +F 3 = P Only 3 of the equations are independent.

3.2. If you specify F, P, W, you can calculate all of the stream variables.

a) Unknown: three stream values F, P, W (plus two compositions if you take into account all of the variables).

b) The two known compositions are not given but may be calculated from Σ x (^) i = 1 ,

c) Two components exist; hence two independent material balances can be written. The problem cannot be solved unless one stream value is specified.

3.2. (a) No. The equations have no solution – they are parallel lines.

The rank of the coefficient matrix is only 1 because the det

⎦⎥^

The rank of the augmented matrix is 2

Largest non zero det. of

⎦⎥^

is of order 2

Thus, although the 2 equations are independent and the number of variables is 2 (the necessary conditions), the sufficient conditions are not met.

Solutions Chapter 3

x 1

x 1 + 2 x 2 = 3

x 2 x 1 + 2 x 2 = 1

(b)

No; 2 solutions The equations are independent

x 2

(x 1 − 1 )^2 + (x 2 − 1 )^2 = 0

x 1 x 1 + x 2 = 1

Two solutions exist as can be seen from the plot hence no unique solution exists.

3.2.

The number of independent equations is just 3. The number of unknown quantities is 3, hence a unique solution is possible.

3.2. No for both

a)

rank of coefficient matrix is2. rank of augmented matrix is2.

r = 2, n = 3 multiple solutions exist

b)

3d column issum of1st two columns,so rank is2. rank of augmented matrix is2, also.

r = 2, n = 3 multiple solutions exist

3.2.

(a) F; (b) F; (c) F (the maximum can be more than the number of independent equations).

3.2.

F = 10 kg P = 16 kg

A = 6 kg

ωF 1 = 0. ωF 2 =? ωF 3 = 0 (assumed)

ωA 1 = 0. ωA 2 =? ωA 3 = 0.

ωP 1 = 0. ωP 2 =? ωP 3 =?

Unknowns (4): (^) ω 2 F^ , ω 2 A , ω 2 P ,ω 3 P

Equations: Mass balances: (1): F(0.10) + A(0.30) = P(0.175) or 10(0.10) + 6(0.30) = 16(0.175) redundant (2): F ( ω 2 F)+ A ( ω 2 A)= P ( ω 2 P) or 10 ( ω 2 F)+6 ( ω 2 A)= 16 ( ω 2 P) (3): F(0) +A(0.20) = P ( ω 2 P) or 10(0) + 6(0.20) = 16 ( ω 2 P)

F A P P 2 ,^2 ,^2 , 3 10 6 -16 0 coeff. matrix 0 0 -0 16

ω ω ω ω

Two independent material balances exist (the rank of the coefficient matrix is 2). In addition three sum of mole fraction equations exist. The total number of equations is 5, hence the degrees of freedom = -1. The problem is overspecified, and will require at least squares solution.

3.2. Examine the row of C 3 H 8. None of the concentrations are greater than the desired 50% so 50% is not achievable by any combination of A, B, or C. Or look at the CH 4 row.

3.2. A B C D Basis: D =100 lb x1 x2 x 5.0 0 0 1. 90.0 10.0 0 31. Coefficient matrix is 5.0 85.0 8.0 53. 0 5.0 80.0 12. 0 0 12.0 14

⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦

M M M M M

No unique solution exists with 5 equations and 4 variables. A least squares solution could be determined, or the 4 equations with the most accurate data solved.

The rank of the coefficient matrix is 3 The rank of the augmented matrix is 4

Hence no unique solution exists.

3.2. You can see by inspection that no combination of tanks 1,2 and 3 will give a mole fraction of 0.52 for the mixture.

Basis: 2.50 mol of tank 4

Let xi = be the total moles of tank i

0.23x 1 + 0.20x 2 + 0.54x 3 = 0.25(2.5) = 0. The balances are 0.36x 1 + 0.33x 2 + 0.27x 3 = 0.23(2.5) = 0. 0.41x 1 + 0.47x 2 + 0.19x 3 = 0.52(2.5) = 1.

The coefficient matrix has a rank of 3 as does the augmented matrix so the set of equations has a solution. However, the solution is

x 1 = −4. x 2 = 6. x 3 = − 0.

The values of x 1 and x 3 are not physically realizable!

Solutions Chapter 3

3.2.

Number of unknowns:

F, W, P and 9 compositions: 12

Equations Specifications: xWCF = 0 1 Sum of mole fractions = 1 3 Material balances (3 species) 3 7

Degrees of freedom = 5

You can make any set of measurements that results in independent equations (assuming equal accuracy). For example, you cannot use all the flows, or all the compositions in one stream, as the resulting set of material balances will not consist of 5 independent balances, but a lesser number.

3.2. The inerts are treated as a compound.

Unknowns: 5 compounds × 5 streams plus 5 streams = 30

Equations: Specifications: Concentrations not shown in diagram assumed to be zero 8 Concentrations with % given 7 E = 11 kg 1 Material balances: 5 5 Implicit equations (^) ( ∑ ω (^) i =1) 5 26

a. Degrees of freedom (additional specifications) = 4

You must make 4 measurements that result in independent equations

b. No.

Solutions Chapter 3

3.2.14a a.

b. The remaining gas is 100% minus the N 2 , and was put on the figure as R.

c. Basis: 100 mol A

d. Unknowns: A, B, C Equations: N 2 and R material balances Basis = A = 100 mol Note: You could treat the values of R as unknowns in each stream, and then there would be 3 more unknowns and 3 more independent equations ( ∑ xi = 1 or ∑m = total mass flow )i i

e. & f. N 2 : 100(0.90) + B(0.30) = C(0.65) R: 100(0.10) + B(0.70) = C(0.35) Total: 100 + B = C

Two of the above equations are independent

g. Solution: B = 71.4 mol C = 171.4 mol

A (^) = 100 = 1. B 71.

Steps 6 and 7: Basis: 2000 lb F Balances: H 2 O Unknowns: P, W Solids

Steps 8 and 9: Total: 2000 = P + W Solids: 2000 (0.30) = P (0.75) P = 800 lb (^) W = 2000 −800 = 1200 lb

4.1.

Step 5: Basis: 1 min Steps 2, 3, 4:

B

D

P

220 mL V 215 mL Urea 2.30 mg/mL H 2 O 220 mL

1.70 mg/mL H 2 O 215 mL

Steps 6 and 7: Unknowns : mP urea and mP H O 2 with two equations: urea and H 2 O. Degrees of freedom = 0. Steps 8 and 9:

(a) H 2 O balance: 220 mL − 215 mL = 5 mL

Urea balance: 2.30 mg 220 mL^ 1.70 mg 215 mL = 141 mg mL mL

−

(b) Dialisate: 1500 + 5 = 1505 mL Urea: 141 mg

Concentration: 141/1505 = 0.0934 mg/mL

4.1. Step 5: Basis: 1 day

Steps 2, 3, 4:

W ton 1.00 H 2 O

P (ton) NaOH^ 0. H 2 O 0.

F=2 ton Mass fr.

NaOH H 2 O

mass fr

Step 6: Unknowns, 2: P, W

Step 7: Balances 2: NaOH, H 2 O, total (2 of the 3 )

Steps 8 & 9:

Total: 2 = W + P NaOH: 2 (.03) = P(.18)

P = 1/3 ton (666 lb)

W = 1 2/3 ton (3334 lb)

4.1. Steps 2, 3, 4: 500 lb 10% solution

3000 lb 13% polymer

A 20% solution

pure solvent S

Step 5: Basis: 3000 lb of final product Steps 6 and 7: Unknowns: A and S; balances: total and polymer

Solutions Chapter 4

Steps 8 and 9:

Total wt A + S + 500 = 3000 A + S = 2500 Polymer 0.2A + 50 = 390

A = 5 (340) = 1700

S = 2500 – 1700 = 800

4.1. Steps 1, 2, 3 and 4:

N = nitrocellulose S – solvent

Treat the problem as a steady state flow problem, or as an unsteady state batch system. As a flow system:

F (lb) N 0. S 0.

Plant P 1000 lb N^ 0. S 0.

Mass fr.

M (lb) N 100%

Step 5: Basis: 1000 lb P

Steps 6 and 7: Two unknowns, F and M. Two balances can be made, N and S.

Steps 8 and 9: Solve to get N: F ( 0. 055 ) + M ( 1. 00 ) = 1000 ( 0. 008 )

S: F ( 9. 045 ) + M ( 0 ) = 1000 ( 0. 92 )

F = 973. 5 lb

M = 26. 5 lb

P = 1000. 0 lb

Solutions Chapter 4

4.1.

Steps 1, 2, 3 and 4:

F = 100 kg mol/min System Mol fr. CH 0. He 0.20 P (kg mol)

W (kg mol)

Mol fr. CH 0. He 0.

Mol.

4

4

n (^) CH 4 nHe

CH 4

He

Step 5: Basis 1 min → 100 kg mol F Step 4: To get P, calculate first the average mol. wt. of F and P, or transform the mole fractions to mass fractions.

Step 5: Basis: 1.00 mol F Mol MW kg CH4 0.80 16.03 12. He 0.20 4 0. 1.00 13. Basis : 1.00 mol P Mol MW kg CH4 0.50 16.03 8. He 0.50 4 2. 1.00 10.

100 kg mol F 13.6 kgF 1.00kg mol F

0.20 kgP 1kgF

1kgmol P 10.01kgP

P = 27.2 kg mol Steps 6, 7, 8 and 9: In W Mol Mol.fr. 100(0.80) = 27.2(0.50) + n (^) CH (^4) 100(0.20) = 27.2(0.50) + n (^) He

n (^) CH (^4) n (^) H (^) e

4.1. Fermenter 40% void 60% cell volume discharge

Separator wet cell 75% H O 25 % dry cell

100% H O

1

2

2

F F

2

W

1

Basis: 1000 cm^3 F (^1)

Drycells = 1000cm 3 in fermenter 60cm 3 cell 100 cm 3 in fermenter

1.1gcell 1cm 3 cell

25gsolids 100g cell

= 16.5gDrycell / 1000cm 3 in fermenter

4.1.

a. Basis: 1 kg mole of mixture

Three components exist: (CH 4 ) (^) x, (C 2 H 6 ) (^) x, (C 3 H 8 ) (^) x

Let A, B, C respectively represent kg mol of each mixture; these are the unknowns.

Equations: 0.25A + 0.35B + 0.55C = 0.30 (CH 4 ) (^) x balance

0.35A + 0.20B + 0.40C = 0.30 (C 2 H 6 ) (^) x balance

0.40A + 0.45B + 0.05C = 0.40 (C 3 H 8 ) (^) x balance

There is a unique solution to the set of equations (in kg mol)

The solution is (^) A = 0.600 B = 0.350 C = 0.

b. It is proposed to prepare the final mixture by blending four different compounds (A, B, C, D); there will still be three equations, but now there will be four unknowns. Since the rank is now less than n, there will be an infinite number of possible blends of the four mixtures. Not required: (An optimization of a revenue function subject to the equations is needed.)

4.1.

a.

CH 4 100% (MW = 16) F =? (lb)

P (lb) =? CH 4 CO 2

mol fr.

0.0204 (average)

100% CO 2 100 lb = 2.27 lb mol

A

Steps 2, 3, 4: 100 lb 1 lb mol = 2.27 lb mol 44 lb

Step 5: Basis 1 min

Step 6: Unknowns: F, P

Step 7: Balances: CH 4 , CO (^2)

Steps 8 and 9: Balances in moles

CH 4 : F (1.00) + A(0) = P (0.9796)

CO 2 : F (0) + 2.27 = P (0.0204) P = 111.41 lb mol

4 4 4

111.27 lb mol 0.9796 lb mol CH 16 lb CH F = = 1746 lb/min (a) 1 lb mol P 1 lb mol CH

b. Redo the problem with a new composition for F:

CH 4 0. CO 2 0.

4 2

CH : F (0.99) + A(0) = P (0.9796)

F = 3520 lb/min CO : F (0.01) + 2.72 = P (0.0204)

error 100 = 50% (b) 3520

Solutions Chapter 4

4.1. Steps 1, 2, 3, and 4:

F(kg)

Mass fr. NH 0 gas 1.

Mass fr. NH 0. gas

3

3

Pipeline

P (kg)

NH 3 72.3 kg/min for 12 min

Step 5: Basis: 1 min

Steps 6 and 7: Two unknown, F and P. Two balances can be made, NH 3 and gas. You can use the total balance as a substitute.

Steps 8 and 9: Total F + 72.3 = P NH 3 F (0) + 72.3 = P (0.00382)

P =18,900 kg / min (^) F = 18,900kg/min or 1.14 ×10 kg/hr^6

Step 10: Check using the gas balance

4.1.

Steps 1, 2, 3, and 4:

10 lb/hr

F

H 2

Br 0

Mass fr.

Mass fr. 1- 0.

H 2 O 1.

P

O

Br

Step 5: Basis: 1 hr

Step 6: Unknowns: F, P

Solutions Chapter 4

Step 7: Balances: H 2 O, Br Steps 8 and 9: Total: F + 10 = P Br: F (0) + 10 = P (0.00012) ; P = 8.33 × 10 4 lb/hr H 2 O: F (1.00) + 0 = P (1-.00012) F = 8.33 × 104 − 10 = 8.33 × 10 4 lb / hr

4.1.

Step 5: Basis: 1 year

Steps 2, 3, 4:

F =? lb P = 15 x10^6 lb

PET PVC

mass fr.

W PVC 100%

(10 ppm)

PET PVC

mass fr.

Steps 6 and 7:

Two unknowns: F, W Two balances: PET, PVC

Steps 8 and 9:

PET balance: F (.98) = P (.999990) ≅ 15 × 106 PVC balance: F (.02) = W (1.00) + P (0.000010) = W + (15 × 106 (10 -5^ ) Total balance could be used in lieu of one of the above

F = 15 × 106 + W

From PET: F ≅^ 15.3 ×^106 lb

W = (15.3 × 106 ) (0.02) – 15 × 106 (10 -5^ ) = 0.31 × 106 lb

H 2 O:

The composition of the original solution is

mass fr. BaNO 3 100g 0. H 2 O 294.1g 0.

Steps 6 and 7:

We have two unknowns, F and C , and can make two independent mass balances so that the problem has a unique solution.

Steps 8 and 9:

Transport through Balance Final solution Initial solution boundary (out)

Ba(NO 3 ) 2 : F(0.0476) - 100 = -C(0.9615) H 2 O: F(0.9524) - 294.1 = -C(0.0385) Total: F - (100 + 294.1) = -C

Solve the Ba(NO 3 ) 2 and total balance to get

F = 305.2 g C = 88.89 g

Step 10: Check using the water balance

305.2(0.9524) – 294.1? -88.89(0.0385)

• 3.42 = -3.

The Ba(NO 3 ) 2 that precipitates out on a dry basis is

3 2 3 2

88.89 g C 0.9615 g Ba(NO ) = 85.5 g Ba(NO ) 1 g C

4.1.

feed = F 60% 1% impurity 39%

H O

W

2

i

S

C10°C

satd soln 1.4 lb

? lb impurity

I

crystals 0.06 lb satd. soln/lb crystals W 0

0.1 % impurity 0%

dried cyrstals

D

II

Na 2 S 2 O 2 = 142 Na 2 S 2 O (^2) ⋅ 5H 2 0 = 232

H 2 O

Na 2 S 2 O 2 ⋅ 5H 2 O 1 lb H 2 O

Na 2 S 2 O 2 ⋅^ 5H 2 O

Na 2 S 2 O 2 ⋅ 5H 2 O

H 2 O as H (^2)

Na S 2 2 O 2

H O 2

M.W.

O

M.W.

Process II Compositions:

a) Stream D Basis: 1 lb mol Na 2 S 2 O 2 ⋅ 5 H 2 O, impurity free

Comp. lb mol mol wt lb wt fr Na 2 S 2 O 2 1 142 142 0. H 2 O 5 18 90 0. Total 6 232 1.

Basis: 100 lb D (Na 2 S 2 O 2 ⋅ 5 H 2 O = 99.9 lb)

Comp. lb Na 2 S 2 O 2 99(0.612) = 61. H 2 O 99(0.388) = 38. impurity 0. Total 100.

Stream C Let y = lb impurity/100 lb free water in saturated solution

Basis: 100 lb dry crystals, impurity free

lb from dry + lb from adhering soln

Comp. crystals from calculation below = Total

Solutions Chapter 4

Na 2 S 2 O 2 61.2 (6)

240 + y

240 + y

H 2 O 38.8 (6)

240 + y

240 + y

impurity ----- (6) y 240 + y

⎟ y 240 + y

Totals 100.0 6 106

Stream S Basis: 100 lb free water in satd. soln., impurity free

Comp. lb from salt lb from free water Total wt. fr. wt.fr.total

Na 2 S 2 O 2 1.4(61.2) = 85.68 85.7 0.

(240 + y )

H 2 O 1.4(38.8) = 54.32 100 154.3 0.

(240 + y) __

140.00 100 240.0 1.

impurity = y 240 + y Total

Basis: 100 lb F

(Rather than use the impurity balance, use the total balance instead if you want)

Water in, W (^) i , comes from balances on Unit I.

Total: 100 + Wi = S + C

Solutions Chapter 4

Na 2 S 2 O 2 60 =

240 + y

⎟ S +

240 + y

C

H 2 O 39 + W (^) i =

240 + y

⎟ S +

240 + y

C

Unknowns: W (^) i ,S,C, y

Total:

Balances on Unit II needed as well because 4 unknowns exist.

Total: C = W 0 + D

Na 2 S 2 O (^2)

240 + y

⎟ C

= 0.611D

H 2 O

240 + y

⎟ C

= W (^) o + 0.388D

Unknowns:W (^) o ,D

Total:

6 equations and 6 unknowns

Note that the complex term involving y can be eliminated to solve for D, W 0 , W (^) i, and S, i.e., in effect making total Na 2 S 2 O 2 and H 2 O balances overall the process.

The solution is (a)W (^) i = 23.34 lb (b)66.5%

4.1. Basis: 100 lb pulp as received

Comp lb = % H 2 O 22 Pulp 78 Freight cost = $1.00 2000 lb= $20.00/ton 100 lb 1 ton Total 100

Assume air dried pulp means the 12% moisture pulp.

Allowed water = 78 lb pulp 12 lb H O^2 = 10.65 lb H O 2 88 lb pulp

Pulp on contract basis = 78 + 10.65 = 88.65 lb

Basis: 1 ton pulp as received

Cost = 88.65 lb 12% pulp rec'd 1 ton shipped $60. $53.19/ton 100 lb shipped 1 ton

4.1. You pay for soap plus transportation.

Basis: 100 kg soap with 30% water

Convert soap in wet soap to soap in dry soap. Data:

W (wet soap) D(dry soap) H 2 O 30 kg 5% soap 70 kg 95% 100 kg 100 Soap balance: 0.70W = 0.95D or 0.70(100) = 0.95D D = 73.68 kg of which 70 kg is soap

Cost of W at your site (containing 70 kg soap): 100 ($0.30 +

Cost of D at your site (containing 70 kg soap): 73. $6. $36. 100

x$ kg

=

x = $ 0.43/kg

4.1.

The problem here is to decide on the balances to make. Not all will be indept balances.

Steps 1, 2, 3, and 4:

F = 100 Mass fr. H (^2) VM C

Mixer

P

H(lb) Mass fr. VM 0. C 0. H O 0.

lb

Ash

(lb) (^2) 1.

O

m m m

m (^) H 2 O VM C Ash

Mass fr. H O 2 V C

M

Ash

VM C

1.00 P

ω

ω

Mass

Step 5: Basis: F = 100 lb

Step 6: Unknowns: mH 2 O , mVM , mC , mAsh , P, H (6)

Step 7: Balances: H 2 O, VM, C, Ash, Σmi = P, mH 2 O /P = 0. mAsh /P = 0.10: (7), and presumably 6 are independent

Steps 8 and 9:

In Out H 2 O (lb): 100 (0.124) + H (0.082) = mH 2 O

VM :(lb) 100 (0.166) + H (0.082) = m (^) VM

C (lb): 100 (0.575) + H (0.887) = m (^) C

Ash (lb) 100 (0.135) + H (0) = mAsh

Total 100 + H = P

Solutions Chapter 4

Solution:

mAsh = 13.5 lb mH 2 O = 13.

Not all of the equations have to be written down as above. Note that ash is a tie element to P, so that the ash balance gives

100(0.135) + H(0) = P(0.10) P = 135 lb and mAsh = 13.5 lb

From a total balance

H + 100 = P H = 35 lb

Step 10: Check via H 2 O balance

12.4 + 35(0.031) = 13. ok 13.49 =13.

4.1.

Step 5: Basis: 1 day

Steps 2, 3, and 4:

D

B

F=

kg mol/day

C 3

i − C (^4) n − C (^4) C 5 +

C 3

i − C (^4) n − C (^4)

i − C n C (^4) C (^5)

Steps 6 and 7:

There are two unknowns, and we have 4 independent equations, but the precision of measurement is not the same in each equation, hence different results are obtained depending on the equations used. Choose the most accurate to use.

Solutions Chapter 4

In Out C 3 (0.019)(5804) = 110 = 0.034D i-C 4 2992 = 0.057D + 0.011B n-C 4 2667 = 0.009D + 0.976B C 5 +^37 = 0.013B Total: 5804 = D + B

Steps 8 and 9:

Use total + i-C 4 balances: kg mol/day 2992 = 0.975(5804-B) + 0.011 B

B = 2563

D = 3240

Use total + n-C 4 balances: 2667 = 0.009 (5804 - B) + 0.976B

B = 2704

D = 3100

The other two balances give

C 3 as a tie component: 5804 kgmol F 1day

0.019kg mol C 3 1.00 kgmol F

1kg mol D 0.034 kg mol C 3

kgmol D day C 5 + as a tie element: 5804 kgmol F 1day

0.006kg mol C 5 1.00 kg mol F

• (^) 1kgmol B 0.013 kgmol C 5 +^ = 2679 kgmol B / day

Two summations exist: ∑ω iA^ = 1 and ∑ω iB^ = 1

hence, we have an adequate number of balances (unless some are not independent) to solve the problem.

Material balance equations in grams

Total: (^1000) +F 1000 +^ S^ ==^ AA^ ++^ BB

NH 3 : 1000(0.0633) + 0 = AωNHA^3 + BωNHB (^3)

N 2 : 1000(0.9367) + 0 = AωNHA^2 + 0

S: (^0) + 1000 = 0 + BωSB

Constraints

ωNHA^3 + ωNA 2 = 1

ωNHB^3 + ωSB^ = 1

ωNHA^3 = 2 ωNH 3B

Solve in Polymath

Note: The equations can be converted to linear equations by letting m (^) NHA^3 = ωNHA^3 A, m (^) NA 2 = ωNA 2 A, etc. The set can be solved in Polymath or reduced to a quadratic equation in m (^) NHB^3 or m (^) NHA^3.

4.1.

Basis: 12 hours

The process is illustrated in the figure

MTBE

H 2 O

H 2 O

MTBE

H 2 O

mass fr. = 1.00 MTBE

The MTBE entering the pond in 12 hours is

3 3

25 boats 0.5 L gasoline 1000 cm 0.72 g 0.10 g MTBE = 900 g MTBE 1 boat I L 1 cm 1 g gasoline

The pond holds (ignoring the MTBE in the pond which is negligible)

3000 m 1,000 m 3m 1000 kg H 2 O 1 m 3 H 2 O

= 9 × 10 9 kg H 2 O

The increase in the concentration of MTBE is

900 g MTBE 9 × 10 9 kg H 2 O

1 kg 1000 g

= 1× 10 -10^ g MTBE/g H 2 O

Solutions Chapter 5

5.1.1 BaCl 2 + Na 2 SO 4 → BaSO 4 + 2NaCl

Mol. wt.: 208.3 142.05 233.4 58.

a. Basis: 5.0 g Na 2 SO (^4)

Example: 5 g Na 2 SO 4 1 g mol Na 2 SO (^4) 142.05 g Na 2 SO (^4)

1 g mol BaCl (^2) 1 g mol Na 2 SO (^4)

208.3 g BaCl (^2) 1 g mol BaCl (^2)

7.33 g BaCl (^2)

Problem Basis Answer b. 5 g BaSo 4 4.47 g BaCl 2 c. 5 g NaCl 8.91 g BaCl 2 d. 5 g BaCl 2 3.41 g Na 2 SO (^4) e. 5 g BaSO 4 3.04 g Na 2 SO (^4) f. 5 lb NaCl 6.08 lb Na 2 SO (^4) g. 5 lb BaCl 2 5.59 lb BaSO 4 h. 5 lb Na 2 SO 4 8.21 lb BaSO 4 i. 5 lb NaCl 9.98 lb BaSO 4

5.1.2 AgNO 3 + NaCl → AgCl + NaNO 3

Mol. wt.: 169.89 58.45 143.3 85.

a. Basis: 5.0 g NaCl

Example: 5.0 NaCl 1 g mol NaCl 58.45 g NaCl

1 g mol AgNO (^3) 1 g mol NaCl

169.89 g AgNO (^3) 1 g mol AgNO (^3)

14.53 g AgNO 3

Problem Basis Answer b. 5 g AgCl 5.92 g AgNO 3 c. 5 g NaNO 3 10.00 g AgNO 3 d. 5 g AgNO 3 1.721 g NaCl e. 5 g AgCl 2.04 g NaCl f. 5 lb NaNO 3 3.44 lb NaCl g. 5 lb AgNO 3 4.22 lb AgCl h. 5 lb NaCl 12.27 lb AgCl i. 5 lb AgNO 3 4.22 lb AgCl

Solutions Chapter 5

5.1.

a. The balanced equation is:

3 As 2 S 3 + 4 H 2 O + 28 HNO 3 = 28NO + 6 H 3 AsO 4 + 9 H 2 SO (^4)

and the reaction is unique within relative proportions.

b. 2 KC1O 3 + 4 H C1 = 2 KC1 + 2 C1O 2 + C1 2 + 2 H 2 O

or KC1O 3 + 6 H C1 = KC1 + 3 C1 2 + 3 H 2 O

or 3 KC1O 3 + 10 HC1 = 3 KC1 + 2 C1O 2 + 4 C1 2 + 5 H 2 O

We classify these reactions as non-unique since they are not simply proportional equations, but are linearly independent, and infinitely many equations can be obtained by linear combination.

5.1.

Basis: 2 g mol

C: 6 × 12 = 72 H: 8 × 1 = 8 O: 6 × 16 = 96 mol. wt. = 72 + 8 + 96 = 176

2 g mol 176 g 1 g mol

1 lb 454 g = 0.775 lb

5.1.

a) LHS RHS

C 4 4 H 10 10 Zn 1 1 O 14 14

Yes, the equation is balanced.

b) Basis: 1.5 kg ZnO

1.5 kg ZnO 1000 g ZnO 1.0 kg ZnO

1 gmol ZnO 81.40 g ZnO

1 gmol DEZ 1 gmol ZnO

123.4 gDEZ 1 gmol DEZ

×

1.0 kg DEZ 1000 g DEZ

= 2.27 kg DEZ

c) 20 cm 3 H 2 O 1.0 g H 2 O 1 cm 3 H 2 O

1.0 gmol H 2 O 18.0 g H 2 O

1 gmol DEZ 5 gmol H 2 O

123.4 g DEZ 1 gmol DEZ

= 27.4 g DEZ

5.1.

Basis: Data given in problem statement

2 3 2 3

55.847 gFe 1 gmol Fe 1 gmol X O 2 gmol X (^) = 1 gmol X 1 55.847 gFe 2 gmol Fe 1 gmol X O 2 3 2 3

55.847 gFe 1 gmol Fe 1 gmol X O 3 gmol O 16 g O = 240 g O 1 55.847 gFe 2 gmol Fe 1 gmol X O 1 gmol O

50.982 g X O 2 3 −24.0 g O = 26.982 g X

26.982 g X g = 26.982 hence X =A 1 gmol X gmol

5.1.

Basis: Data given in problem statement

2 2 2 2

1.603g CO 1 gmol CO 1 gmol C 12.00g C = 0.437 g C 1 44.01g CO 1 gmol CO 1 gmol C 2 2 2 2

0.2810gH O 1 gmol H O 2 gmol H 1.008g H = 0.03144 g H 1 18.02g H O 1 gmol H O 1 gmol H

0.6349 g Cx H (^) y O (^) z – 0.437g C – 0.03144g H = 0.1661g O

Based on O:

0.1661g O 1 gmol O (^) = 0.01038 gmol O= 1 gmol O 1 16.00g O 0. 0.437 g C 1 gmol C 0.03642 gmol C = = 3.5 gmol C 1 12.00 g C 0. 0.03144 g H 1 gmol H (^) = 0.03119 gmol H= 3 gmol H 1 1.008 g H 0.

The formula in integers is

C 7 H 6 O (^2)

5.1.

Basis: Data given in the problem

Mass of hydrate 10.407 g Mass of dry sample -9.520 g Mass of water 0.887 g

(^2 ) 2

9.520 g BaI 1 g mol BaI (^) = 0.0243 g mol BaI 391 g BaI

(^2 ) 2

0.887 g H O 1 g mol H O = 0.0493 g mol H O 18.02 g H O

2 2 2 2

0.0243 = 0.49, or 2 H O for 1 BaI. Thus the hydrate is BaI 2H O.

⋅

Solutions Chapter 5

5.1.

Basis: 2000 tons 93.2% H 2 SO 4 (1 day)

a. 2000 tons soln 0.932 ton H

2 SO 4

1 ton soln

1 ton mole H 2 SO (^4) 98 ton H 2 SO (^4)

1 mol S 1 mol H 2 SO (^4)

32 ton S 1 ton mol S

= (^) 609 ton S

b. 609 ton S^ 1 ton mol S 32 ton S

1.5 mol O (^2) 1 mol S

32 ton O (^2) 1 ton mol O (^2) = 913.5 ton O 2

c. 609 ton S^ 1 ton mole S 32 ton S

1 mole H 2 O 1 mol S

18 ton H 2 O 1 ton mol H 2 O = 342.6 ton H 2 O

5.1.

Basis: 1 lb Br 2 2Br −^ →1Br 2

2Br–^ + Cl 2 → 2Cl–^ + Br 2 MW: 70.9 159.

Br 2 + C 2 H 4 → C 2 H 4 Br MW: 28 187.

a. 0.27 lb acid 2000 lb seawater

1 × 10 6 lb seawater 65 lb Br (^2) = 2.08 lb 98% H 2 SO 4 / lb Br (^2)

b.

1 lb Br 2 mole Br (^2) 159.9 lb Br (^2)

1 mole Cl (^2) 1 mole Br (^2)

70.9 lb Cl (^2) mole Cl (^2) = 0.445 lb Cl (^2)

c. 1 lb Br 2 1 × 10 6 lb sea water 65 lb Br (^2) = 15,400 lb seawater

(^2 2 2 4 2 2 4 22 4 ) 2 2 2 4 2

1 lb Br mol Br 1 mol C H Br 187.9 lb C H Br d. = 1.176 lb C H Br 159.9 lb Br 1 mol Br 1 mol C H Br

Solutions Chapter 5

5.1. MW MW FeSO 4 ⋅7H 2 O 277.9 FeSO 4 ⋅H 2 O 169.

FeSO 4 ⋅4H 2 O 223.9 FeSO 4 151.

It is best to evaluate the three types of ferrous sulfate in terms of the cost/ton of FeSO 4

Basis: 475 ton material a. (^) FeSO 4 ⋅7H 2 O:

$83,766. 475 ton FeSO 4 ⋅ 7H 2 O = $176.35/ton FeSO 4 ⋅ 7H O 2 , which is equivalent to

1 ton FeSO 4 ⋅ 7H 2 O

277.9 ton FeSO 4 ⋅ 7H 2 O 1 ton mol FeSO 4 ⋅ 7H 2 O

1 ton mol FeSO 4 ⋅7H 2 O 1 ton mol FeSO 4

1 ton mol FeSO (^4) 151.9 ton FeSO (^4)

= $323/ton FeSO 4

b. FeSO 4 ⋅4H 2 O:

$ ton FeSO (^4)

151.9 ton FeSO (^4) 1 ton mol FeSO 4

1 ton mol FeSO (^4) 1 ton mol FeSO 4 ⋅ 4H 2 O

1 ton mol FeSO 4 ⋅ 4H 2 O 223.9 ton FeSO 4 ⋅ 4H 2 O

= $219 / ton FeSO 4 ⋅ 4H 2 O

c. (^) FeSO 4 ⋅H 2 O:

$ ton FeSO (^4)

151.9 ton FeSO (^4) 1 ton mol FeSO 4

1 ton mol FeSO (^4) 1 ton mol FeSO 4 ⋅ H 2 O

1 ton mol FeSO 4 ⋅H 2 O 169.9 ton FeSO 4 ⋅ H 2 O

= $289 / ton FeSO 4 ⋅ H 2 O