1) Two methanol-water mixtures are contained in separate flasks. The mixture
contains 40 wt% methanols, and the second contains 70 wt% methanols. If 200 g
of the first mixture are combined with 150 g of the second, what are the mass and
composition of the product?
Solution:
m1m = 0.4
m1w = 0.6
Flask
1
200 gm
Products
Combination
m2m = 0.7
m2w = 0.3
Flask
2
150 gm
- assume steady state
- no chemical reactions
INPUT = OUTPUT
mTm (IN) = 0.4 x 200 + 0.7 x 150 = 185 gm = mTm (OUT)
mTw (IN) = 0.6 x 200 + 0.3 x 150 = 165 gm = mTw (OUT)
mTp = mTm + mTw = 185 + 165 = 350gm
methane = 52.86 % by mass
water = 47.14 % by mass
2) One thousand Kilograms per hour of a mixture containing equal parts by mass
of benzene (C 6H6) and toluene (C 7H8) are distilled. The flow rate of the overhead
product stream is 488 Kg/h, and the bottom stream contains 7.11 wt % benzene.
Draw and label a flowchart for the process. Then calculate the mass and mole
fractions of benzene and the molar flow rates of benzene and toluene (mole/h) in
the overhead product stream.
OPo = 488 kg/h
m•bo = ?
m•to = ?
Solution:
a)
F = 1000 kg/h
m•b = 500 kg/h
m•t = 500 kg/h
Distillation
OPl = ?
m•bl = 0.0711 OPl
m•tl = ?
CMB:
F = OPo + OPl
;
OPl = F - OPo = 1000 – 488 = 512 kg/h
m•bl = 0.0711 OPl = 0.0711 x 512 = 36.4 kg/h
m•b = m•bl + m•bo
;
m•bo = m•b - m•bl = 463.6 kg.h
m•to = OPo - m•bo = 488 – 463.3 = 24.4 kg/h
m•tl = m•t - m•to = 500 – 24.4 = 475.6 kg/h
Check:
m•tl + m•bl = 512
475.6 + 36.4 = 512
b) mass and mole fractions of benzene:
Mwtb = 78 gm/mol
Mwtl = 92 gm/mol
Xb = mb / mT = 0.5
Xbo = 463.6 / 488 = 0.95
Xbl = 36.4 / 512 = 0.0711
nb = 500 / 78 = 6.4 mol
nt = 500 / 92 = 5.4 mol
Yb = 5.4 / 11.8 = 0.46
nT = 11.8 mol
nbo = 463.6 / 78 = 5.9 mol
nto = 24.4 / 92 = 0.27 mol
Ybo = 5.9 / 6.2 = 0.95
nTo = 6.2 mol
nbl = 36.4 / 78 = 0.47 mol
ntl = 475.6 / 92 = 5.17 mol
Ybl = 0.47 / 5.64 = 0.08
nTl = 5.64
c) molar flow rate of benzene and toluene in the overhead stream:
m•bo = 463.6 kg /h
n•bo = 463.6 / 78 = 5.9 kg/h
m•to = 24.4 kg/h
n•to = 24.4 / 92 = 0.27 kg/h
3) Strawberries contain 15 % solids and water 85 % by mass. To make strawberry
jam, crushed strawberries and sugar are mixed in a 45: 55 ratio, and the mixture is
heated to evaporate water until the residue contains 1/ 3 water by mass. Draw an d
label a flowchart of this process, and use it to calculate how many Kgs of
strawberries are needed to make 1 Kg of jam.
Solution:
Evep Water
m (Str) = 45 gm
m (W.e) = ?
Mixer
Heater
m (Su) = 55 gm
Jam
IN
m (wi) = 0.85 x 45 = 38.25 gm
m (w) = 1/3 J
m (s) = ?
m (su) = ?
OUT
m (si) = 0.15 x 45 = 6.75 gm
CMB :
Suger: In = Out
m (sui) = 55 gm = m (suo)
Solides : In = Out
m (si) = 6.75 gm = m (so)
m (so) + m (so) = 2/3 m (J) = 61.75 gm
m (J) = 92.625 gm
m(wo) = 1/3 m (J) = 30.875 gm
m (wi) = m (we) + m(wo)
m (we) = 38.25 – 30.875 = 7.375 gm
J
0.092625 kg
Str
0.045 kg
1 kg
x =?
x = 0.486 kg
Answer:
0.486 kgs of strawberry are necessary to make 1 kg jam.
Appendix
Used Units and their conversions:
 Volume [V] in :
1 m3
= 1000 L
1L
= 1000 cm3
1 m3
= 106 cm3

Number of Moles [n] in :
mol

Molecular weight [Mwt] in:
g/mol

Mass [m] in:
1kg
= 1000 gm