ChE 110 LE 1 Key to Correction
I. Multiple Choice (2 points each). Provide the needed information. Write your answers on the space
provided.
1. Dry air requirement (in ft3) for burning 1 ft3 of CO to CO2 may be around
(a) 2.4
(b) 0.87
(c) 1.75
(d) 11.4
1
𝐶𝑂 + 𝑂2 → 𝐶𝑂2
2
Solution:
1 1 𝑚𝑜𝑙 𝐴𝑖𝑟
𝐷𝑟𝑦 𝑎𝑖𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡 = 1 𝑓𝑡 3 ( ) ( ) = 2.38 𝑓𝑡 3
2 0.21 𝑚𝑜𝑙 𝑂2
2. It is the ratio of the moles of desired product formed to the moles of undesired product formed in a
particular reaction.
(a) Selectivity
(b) Fractional conversion
(c) Extent of reaction
(d) Degree of completion
3. It is the ratio of the change in the number of moles of a particular species and its stoichiometric coefficient
in the reaction.
(a) Fractional yield
(b) Degree of completion
(c) Extent of reaction
(d) Excess reactant
4. A 0.45-lb gaseous mixture contains 20 g of nitrogen gas, 83 g of oxygen gas, and the rest is sulfur
dioxide. Calculate the average molecular weight of the mixture in lb/lbmol.
(a) 45
(b) 42
(c) 38
(d) 21
5. An act establishing the Bicol University, defining its powers, functions, and duties, appropriating funds
therefor, and for other purposes.
(a) RA 5513
(b) RA 5252
(c) RA 5521
(d) RA 5113
6. A large storage tank contains oil having a density of 917 kg/m 3. The tank is 3.66 m tall and is vented
(open) to the atmosphere of 1 atm absolute at the top. The tank is filled with oil to a depth of 3.2 m and
also contains 0.46 m of water in the bottom of the tank. Calculate the pressure in pound force per square
in absolute (psia) at the bottom of the tank.
, II. Problem Solving (5 points each).
1. A solution of ammonium sulfate weighing 15,000 kg with 30 wt% ammonium sulfate is cooled at 293 K.
The salt crystallizes as the decahydrate. Assuming that 3% of the total water in the feed is lost by
evaporation of water in cooling, calculate the yield of hydrated crystals (in kg) if the concentration of the
saturated solution left (called mother liquor) at 293 K is 21.5 kg anhydrous ammonium sulfate per one
hundred kg of total water.
Water (W) = 0.03*(Total water)
Feed (F) = 15,000 kg
30% (NH4)2SO4
Crystallizer
Crystals (C) = (NH4)2SO410H2O
Mother Liquor (ML)
21.5 kg (NH4)2SO4 /100 kg H2O or
21.5 kg (NH4)2SO4/121.5 kg Solution
Solution:
Basis: 15,000 kg Feed
Total water content in the feed: 15,000 kg (0.70) = 10,500 kg
Water lost during cooling: 10,500 kg (0.03) = 315 kg
OMB:
𝐹 = 𝑊 + 𝐶 + 𝑀𝐿
15,000𝑘𝑔 = 315𝑘𝑔 + 𝐶 + 𝑀𝐿
(NH4)2SO4 Balance:
21.5 𝑘𝑔 132 𝑘𝑔
15,000 𝑘𝑔(0.30) = 𝑀𝐿 ( ) + 𝐶( )
121.5 𝑘𝑔 312 𝑘𝑔
𝐶 = 7725 𝑘𝑔 𝑎𝑚𝑚𝑜𝑛𝑖𝑢𝑚 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠
𝐶 = 6959 𝑘𝑔 𝑚𝑜𝑡ℎ𝑒𝑟 𝑙𝑖𝑞𝑢𝑜𝑟
ANS: 7,725 kg
2. Formaldehyde (CH2O) is made by the catalytic oxidation of pure methanol vapor and air in a reactor.
The moles from this reactor (after the reaction) in equilibrium are 63.1 nitrogen gas, 13.4 oxygen gas,
5.9 water, 5.1 formaldehyde, 12.3 methanol, and 1.2 formic acid.
I. Multiple Choice (2 points each). Provide the needed information. Write your answers on the space
provided.
1. Dry air requirement (in ft3) for burning 1 ft3 of CO to CO2 may be around
(a) 2.4
(b) 0.87
(c) 1.75
(d) 11.4
1
𝐶𝑂 + 𝑂2 → 𝐶𝑂2
2
Solution:
1 1 𝑚𝑜𝑙 𝐴𝑖𝑟
𝐷𝑟𝑦 𝑎𝑖𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡 = 1 𝑓𝑡 3 ( ) ( ) = 2.38 𝑓𝑡 3
2 0.21 𝑚𝑜𝑙 𝑂2
2. It is the ratio of the moles of desired product formed to the moles of undesired product formed in a
particular reaction.
(a) Selectivity
(b) Fractional conversion
(c) Extent of reaction
(d) Degree of completion
3. It is the ratio of the change in the number of moles of a particular species and its stoichiometric coefficient
in the reaction.
(a) Fractional yield
(b) Degree of completion
(c) Extent of reaction
(d) Excess reactant
4. A 0.45-lb gaseous mixture contains 20 g of nitrogen gas, 83 g of oxygen gas, and the rest is sulfur
dioxide. Calculate the average molecular weight of the mixture in lb/lbmol.
(a) 45
(b) 42
(c) 38
(d) 21
5. An act establishing the Bicol University, defining its powers, functions, and duties, appropriating funds
therefor, and for other purposes.
(a) RA 5513
(b) RA 5252
(c) RA 5521
(d) RA 5113
6. A large storage tank contains oil having a density of 917 kg/m 3. The tank is 3.66 m tall and is vented
(open) to the atmosphere of 1 atm absolute at the top. The tank is filled with oil to a depth of 3.2 m and
also contains 0.46 m of water in the bottom of the tank. Calculate the pressure in pound force per square
in absolute (psia) at the bottom of the tank.
, II. Problem Solving (5 points each).
1. A solution of ammonium sulfate weighing 15,000 kg with 30 wt% ammonium sulfate is cooled at 293 K.
The salt crystallizes as the decahydrate. Assuming that 3% of the total water in the feed is lost by
evaporation of water in cooling, calculate the yield of hydrated crystals (in kg) if the concentration of the
saturated solution left (called mother liquor) at 293 K is 21.5 kg anhydrous ammonium sulfate per one
hundred kg of total water.
Water (W) = 0.03*(Total water)
Feed (F) = 15,000 kg
30% (NH4)2SO4
Crystallizer
Crystals (C) = (NH4)2SO410H2O
Mother Liquor (ML)
21.5 kg (NH4)2SO4 /100 kg H2O or
21.5 kg (NH4)2SO4/121.5 kg Solution
Solution:
Basis: 15,000 kg Feed
Total water content in the feed: 15,000 kg (0.70) = 10,500 kg
Water lost during cooling: 10,500 kg (0.03) = 315 kg
OMB:
𝐹 = 𝑊 + 𝐶 + 𝑀𝐿
15,000𝑘𝑔 = 315𝑘𝑔 + 𝐶 + 𝑀𝐿
(NH4)2SO4 Balance:
21.5 𝑘𝑔 132 𝑘𝑔
15,000 𝑘𝑔(0.30) = 𝑀𝐿 ( ) + 𝐶( )
121.5 𝑘𝑔 312 𝑘𝑔
𝐶 = 7725 𝑘𝑔 𝑎𝑚𝑚𝑜𝑛𝑖𝑢𝑚 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠
𝐶 = 6959 𝑘𝑔 𝑚𝑜𝑡ℎ𝑒𝑟 𝑙𝑖𝑞𝑢𝑜𝑟
ANS: 7,725 kg
2. Formaldehyde (CH2O) is made by the catalytic oxidation of pure methanol vapor and air in a reactor.
The moles from this reactor (after the reaction) in equilibrium are 63.1 nitrogen gas, 13.4 oxygen gas,
5.9 water, 5.1 formaldehyde, 12.3 methanol, and 1.2 formic acid.
