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Multiple choice question for engineering

Set 1

1. An aqueous solution with sulfur 10 g/L at the rate is 100 L/min and an organic compound with no sulfur at the rate 50 L/min were put into an extraction machine and produced aqueous solution with sulfur 1 g/L, what is the amount of sulfur in organic compound after extraction?
a) 2 g/L
b) 5 g/L
c) 9 g/L
d) 15 g/L

View Answer

Answer: c [Reason:] Let the amount of sulfur in organic compound be x. Sulfur balance equation, 100(10) + 50(0) = 100(1) + 50(x), => x = 9 g/L.

2. An aqueous solution with chlorine 30 g/L and an organic compound with no chlorine at the rate 50 L/min were put into an extraction machine and produced aqueous solution with chlorine 5 g/L and organic compound with chlorine 10 g/L, what is the rate of aqueous solution?
a) 10 L/min
b) 20 L/min
c) 30 L/min
d) 40 L/min

View Answer

Answer: c [Reason:] Let the rate of aqueous solution be x. Chlorine balance equation, x(30) + 50(0) = x(5) + 50(10), => x = 20 L/min.

3. A gas mixture input is given to a membrane with 40% O2 and 60% N2, the waste contains 80% of the input and the product contains 25% O2 and 75% N2, what is the percentage of O2 in the waste?
a) 0.22
b) 0.44
c) 0.66
d) 0.88

View Answer

Answer: b [Reason:] Let the input is 100 Kg mol, => waste = 0.8(100) = 80 Kg mol, => Product = 20 Kg mol. Equation of material balance, O2: 0.4(100) = 0.25(20) + y(80), => y = 0.4375 ≈ 0.44.

4. A gas mixture input is given to a membrane with 20% O2 and 80% N2, the waste contains 60% of the input and the product contains 30% O2 and 70% N2, what are the number of moles of O2 in the waste?
a) 1
b) 2
c) 3
d) Can’t be determined

View Answer

Answer: d [Reason:] Since the amount of input is not given, number of moles of O2 cannot be determined.

5. 100 Kg mol of a gas mixture input is given to a membrane with 30% O2 and 70% N2, the waste contains 60% of the input and the product contains 10% O2 and 90% N2, what are the number of moles of O2 in the waste?
a) 12
b) 16
c) 34
d) 42

View Answer

Answer: c [Reason:] Let the moles of O2 in the waste be x, Amount of waste = 100(0.6) = 60 Kg mol, => amount of product = 40 Kg mol. Material balance equation for O2: 0.3(100) + 0.1(40) = x, => x = 34 moles.

6. A liquid solid mixture with 10% solid and 90% liquid is input to a dryer, some of the liquid evaporated and the product contains 60% solid and 40% liquid, what is the ratio of amount of water evaporated and amount of input?
a) 2:1
b) 3:2
c) 5:4
d) 6:5

View Answer

Answer: d [Reason:] Let the amount of input be F and the amount of evaporated water be W. Solid balance: 0.1*F = 0*W + 0.6*(F – W), => F/W = 6/5.

7. A liquid solid mixture with 20% solid and 80% liquid is input to a dryer, if the amount of evaporated water is 60% of the amount of input, what is the percentage of solid in the product?
a) 40%
b) 50%
c) 80%
d) 90%

View Answer

Answer: b [Reason:] Let the amount of input be F, => the amount of evaporated water be 0.6F, => amount of product = 0.4F. Solid Balance: 0.2(F) = 0(0.6F) + x(0.4F), => x = 0.5, => percentage of solid in the product = 50%.

8. A 100 Kg liquid solid mixture with 10% solid and 90% liquid is input to a dryer, if the amount of evaporated water is 60% of the amount of input, what is the amount of liquid in the product?
a) 10 Kg
b) 20 Kg
c) 30 Kg
d) 40 Kg

View Answer

Answer: c [Reason:] Let the amount of liquid in the product be x. Liquid balance: 0.9(100) = 1(60) + x, => x = 30 Kg.

9. 50 Kg of a solid liquid mixture containing 10% solid and 90% water is left open in atmosphere, after some time the water is 80%, what is the weight of the mixture now?
a) 10 Kg
b) 25 Kg
c) 35 Kg
d) 50 Kg

View Answer

Answer: b [Reason:] Let the new weight of mixture be P. Solid balance: 0.1(50) = 0.2P, => P = 25 Kg.

10. 50 Kg of a solid liquid mixture containing 20% solid and 80% water is left open in atmosphere, after some time the water is 60%, how much water is evaporated?
a) 15 Kg
b) 25 Kg
c) 35 Kg
d) 45 Kg

View Answer

Answer: b [Reason:] Let the weight of the water evaporated be W. Water Balance: 0.8(50) = 1(W) + 0.6(50 – W), => W = 25 Kg.

11. A gas mixture of 40% He and 60% Ne is passed through a diffusion tube and the product has 10% He and 90% Ne, what is the percentage of He recovered?
a) 25%
b) 33.3%
c) 50%
d) 66.6%

View Answer

Answer: b [Reason:] Let the amount of input gas mixture be F, amount of He recovered be D. He balance: 0.4F = 1D + 0.1(F – D), => D/F = 0.333, => percentage of He recovered = 33.3%.

12. 100 Kg of a gas mixture of 40% He and 60% Ne is passed through a diffusion tube and the product has 20% He and 80% Ne, what is the amount of He recovered?
a) 25 Kg
b) 45 Kg
c) 50 Kg
d) 75 Kg

View Answer

Answer: a [Reason:] Let the amount of He recovered be D, He Balance: 0.4(100) = 1(D) + 0.2(100 – D), => D = 25 Kg.

13. A gas mixture of 40% He and 60% Ne is passed through a diffusion tube, if the amount of He is recovered is 20% of the input, what is the percentage of He in the product?
a) 25 %
b) 33.3 %
c) 50 %
d) 66.6 %

View Answer

Answer: a [Reason:] Let the amount of input be F, He balance: 0.4(F) = 1(0.2F) + x(F – 0.2F), => x = 0.25, => Percentage of He in the product = 25%.

14. 400 g of CaSO4 is dissolved with 500 g of H2O, if 226 g of CaSO4.5H2O crystallizes out, then what is the percentage of CaSO4 in the remaining solution?
a) 39.16%
b) 51.54%
c) 62.28%
d) 75.67%

View Answer

Answer: a [Reason:] Moles of crystal = 226/226 = 1, => the crystal has 136 g of CaSO4 and 90 g of water, => Remaining solution has 264 g of CaSO4 and 410 g of H2O, => Percentage of CaSO4 in remaining solution = 264/(264 + 410) *100 = 39.16%.

15. How much nitrogen must be added to a 20% nitrogen solution to obtain 100 Kg of 40% nitrogen solution?
a) 10 Kg
b) 25 Kg
c) 50 Kg
d) 75 Kg

View Answer

Answer: b [Reason:] Let the amount of nitrogen added be x, and amount of 20% nitrogen solution be P. Nitrogen balance: x + 0.2(P) = 0.4(100), Water balance: 0.8(P) = 0.6(100), => P = 75 Kg, => x = 0.4(100) – 0.2(75), => x = 25 Kg.

Set 2

1. The general material balance equation for reactive system is Accumulation = Input + Generation – x, what is x?
a) Consumption – Output
b) Consumption + Output
c) Output – Consumption
d) None of the mentioned

View Answer

Answer: b [Reason:] Accumulation = Input – Output + Generation – Consumption, => x = Consumption + Output.

2. In a system if, Accumulation = Input – Output + Generation – Consumption, where all the four terms are distinct and non-zero, then the system is which of the following?
a) Steady-state
b) Closed
c) Non-reactive
d) None of the mentioned

View Answer

Answer: d [Reason:] As the formula involves input and output it is an open system, and also as it involves generation and consumption the system is reactive, and since the input is not equal to output so the system is unsteady-state, and the answer is d.

3. In a system if, Accumulation = Generation – Consumption, where both the terms are distinct and non-zero, then the system is which of the following?
a) Steady-state
b) Closed
c) Non-reactive
d) None of the mentioned

View Answer

Answer: b [Reason:] As the formula does not involves input and output it is a closed system.

4. In a system if, Accumulation = Input – Output, where both the terms are distinct and non-zero, then the system is which of the following?
a) Steady-state
b) Closed
c) Non-reactive
d) None of the mentioned

View Answer

Answer: c [Reason:] As the formula does not involves generation and consumption it is a non-reactive system.

5. In a steady-state reactive system, NaOH and HCl are supplied to the system both at the rate of 100 mole/min, then what is the rate of NaCl produced?
a) 100 mole/min
b) 200 mole/min
c) 300 mole/min
d) 400 mole/min

View Answer

Answer: b [Reason:] As the system is in steady-state, rate of product = rate of feed, => rate of NaCl = 100 + 100 = 200 mole/min.

6. In a steady-state reactive system, 1 molar each of NaOH and HCl are supplied to the system both at the rate of 100 L/min, then what is the amount of NaCl produced?
a) 1 molar
b) 2 molar
c) 3 molar
d) 4 molar

View Answer

Answer: a [Reason:] The balanced reaction of the system is NaOH + HCl -> NaCl + H2O, => amount of NaCl = amount of HCl = amount of NaOH = 1 molar.

7. In a steady-state reactive system, NaCl and H2CO3 are supplied to the system at the rate of 200 L/min and 100 L/min respectively, what is the rate at which Na2CO3 is produced?
a) 100 L/min
b) 200 L/min
c) 300 L/min
d) 400 L/min

View Answer

Answer: c [Reason:] As the system is in steady state, => rate of product = rate of feed, => rate of Na2CO3 = 200 + 100 = 300 L/min.

8. In a steady-state reactive system, 2 molar NaCl and 1 molar H2CO3 are supplied to the system both at the rate of 100 mole/min, then how much Na2CO3 produced?
a) 1 molar
b) 2 molar
c) 3 molar
d) 4 molar

View Answer

Answer: a [Reason:] The balanced reaction of the system is 2NaCl + H2CO3 -> Na2CO3 + 2HCl, => amount of Na2CO3 = 1 molar.

9. In a steady-state reactive system, 5 molar CH4 and 5 molar O2 are supplied to the system, then what is the amount of CO2 produced?
a) 2.5 molar
b) 5 molar
c) 7.5 molar
d) 10 molar

View Answer

Answer: b [Reason:] The balanced chemical reaction of the system is CH4 + O2 -> CO2 + 2H2O, => amount of CO2 = 5 molar.

10. In a steady-state reactive system, 5 molar KNO3 and 5 molar H2SO4 are supplied to the system, then what is the amount of K2SO4 produced?
a) 2.5 molar
b) 5 molar
c) 7.5 molar
d) 10 molar

View Answer

Answer: a [Reason:] The balanced chemical reaction of the system is 2KNO3 + H2SO4 -> K2SO4 + 2HNO3, => amount of K2SO4 = 2.5 molar.

11. The flow in rate of HCl and NaOH in a system are 1 mole/hr each what is the flow out rate of NaCl?
a) 1 mole/hr
b) 2 mole/hr
c) 3 mole/hr
d) 4 mole/hr

View Answer

Answer: a [Reason:] The reaction is HCl + NaOH -> NaCl + H2O, => flow out rate of NaCl = 1 mole/hr.

12. The flow in rate of H2SO4 and NaCl in a system are 2 mole/hr each what is the flow out rate of HCl?
a) 1 mole/hr
b) 2 mole/hr
c) 3 mole/hr
d) 4 mole/hr

View Answer

Answer: b [Reason:] The reaction is H2SO4 + 2NaCl -> 2HCl + Na2SO4, => flow out rate of HCl = 2*1 = 2 mole/hr.

13. The flow in rate of CH4 and O2 in a system are 2 mole/hr each what is the flow out rate of CO2?
a) 1 mole/hr
b) 2 mole/hr
c) 3 mole/hr
d) 4 mole/hr

View Answer

Answer: a [Reason:] The reaction is CH4 + 2O2 -> CO2 + 2H2O, => flow out rate of CO2 = 1*1 = 1 mole/hr.

14. The flow in rate of C3H8 and O2 in a system are 2 mole/hr each what is the flow out rate of CO2?
a) 1 mole/hr
b) 1.2 mole/hr
c) 1.5 mole/hr
d) 2 mole/hr

View Answer

Answer: b [Reason:] The reaction is C3H8 + 5O2 -> 3CO2 + 4H2O, => flow out rate of CO2 = 2*3/5 = 1.2 mole/hr.

15. The flow in rate of C6H6 and H2 in a system are 3 mole/hr each what is the flow out rate of C6H12?
a) 1 mole/hr
b) 1.2 mole/hr
c) 1.5 mole/hr
d) 2 mole/hr

View Answer

Answer: a [Reason:] The reaction is C6H6 + 3H2 -> C6H12, => flow out rate of C6H12 = 3*1/3 = 1 mole/hr.

Set 3

1. What is the partial pressure of a 40% gas in a system with vapor pressure 1 mm Hg?
a) 0.2 mm Hg
b) 0.4 mm Hg
c) 0.6 mm Hg
d) 0.8 mm Hg

View Answer

Answer: b [Reason:] p = 0.4*1 = 0.4 mm Hg.

2. What is the vapor pressure of 50% gas in a system with partial pressure 2 mm Hg?
a) 1 mm Hg
b) 2 mm Hg
c) 4 mm Hg
d) 8 mm Hg

View Answer

Answer: c [Reason:] P = 2/0.5 = 4 mm Hg.

3. What is the fraction of gas in a system with partial pressure 2 mm Hg and vapor pressure 5 mm Hg?
a) 0.4
b) 0.5
c) 0.6
d) 0.8

View Answer

Answer: a [Reason:] Fraction of gas = 2/5 = 0.4.

4. What is the partial pressure of 20% gas in a system with vapor pressure 10 mm Hg?
a) 2 mm Hg
b) 10 mm Hg
c) 50 mm Hg
d) 100 mm Hg

View Answer

Answer: a [Reason:] Partial pressure = 0.2*10 = 2 mm Hg.

5. What is the vapor pressure of 40% gas in a system with partial pressure 20 mm Hg?
a) 5 mm Hg
b) 10 mm Hg
c) 20 mm Hg
d) 50 mm Hg

View Answer

Answer: d [Reason:] Vapor pressure = 20/0.4 = 50 mm Hg.

6. According to Henry’s Law, what is the fraction of a gas with partial pressure p and Henry’s constant H?
a) p*H
b) p/H
c) p + H
d) p – H

View Answer

Answer: b [Reason:] According to Henry’s Law, p = H*x, => x = p/H.

7. What is the partial pressure of a 10% gas in a system, if H = 2 mm Hg?
a) 0.1 mm Hg
b) 0.2 mm Hg
c) 0.3 mm Hg
d) 0.4 mm Hg

View Answer

Answer: b [Reason:] p = 0.1*2 = 0.2 mm Hg.

8. What is the fraction of a gas in a system with partial pressure 10 mm Hg, H = 25 mm Hg?
a) 0.1
b) 0.2
c) 0.3
d) 0.4

View Answer

Answer: d [Reason:] Fraction of a gas = 10/25 = 0.4.

9. A 40% gas in a system has partial pressure 4 atm, what is Henry’s constant?
a) 1.6 atm
b) 4 atm
c) 10 atm
d) 16 atm

View Answer

Answer: c [Reason:] H = 4/0.4 = 10 atm.

10. What is the partial pressure of a 70% gas in a system, if H = 5 mm Hg?
a) 3.5 mm Hg
b) 7 mm Hg
c) 14 mm Hg
d) 35 mm Hg

View Answer

Answer: a [Reason:] Partial pressure = 0.7*5 = 3.5 mm Hg.

11. What is the K value of a gas with partial pressure 10 atm and total pressure of the system is 50 atm?
a) 0.1
b) 0.2
c) 100
d) 500

View Answer

Answer: b [Reason:] K = 10/50 = 0.2.

12. What is the K value of a gas with partial pressure 50 Pa and total pressure of the system is 1 mm Hg?
a) 0.37
b) 0.45
c) 0.79
d) 0.91

View Answer

Answer: a [Reason:] K = 50/133.3 = 0.37.

13. What is the total pressure of a system having 40% O2 with 5 atm partial pressure and 60% N2 with 10 atm partial pressure?
a) 2 atm
b) 6 atm
c) 8 atm
d) 10 atm

View Answer

Answer: c [Reason:] P = 0.4*5 + 0.6*10 = 8 atm.

14. What is the total pressure of a system having 25% CO2 with 4 atm partial pressure and 30% O2 with 10 atm partial pressure, and 45% SO2 with partial pressure 20 atm?
a) 9 atm
b) 13 atm
c) 17 atm
d) 19 atm

View Answer

Answer: b [Reason:] P = 0.25*4 + 0.3*10 + 0.45*20 = 1 + 3 + 9 = 13 atm.

15. What is the total pressure of a system having 50% H2CO3 with 6 atm partial pressure and 50% O2 with 12 atm partial pressure?
a) 3 atm
b) 6 atm
c) 9 atm
d) 12 atm

View Answer

Answer: c [Reason:] P = 0.5*6 + 0.5*12 = 9 atm.

Set 4

1. Which of the following is the humidity chart?
a) Psychrometric chart
b) Psychometric chart
c) Psychological chart
d) None of the mentioned

View Answer

Answer: a [Reason:] Humidity chart is also called Psychrometric chart.

2. Humidity chart relates which of the following balances?
a) Momentum and energy balance
b) Material and energy balance
c) Material and momentum balance
d) None of the mentioned

View Answer

Answer: b [Reason:] Humidity chart relates material and energy balance.

3. In the humidity chart what does the horizontal axis represent?
a) Air temperature
b) Dry-Bulb temperature
c) Wet-Bulb temperature
d) None of the mentioned

View Answer

Answer: b [Reason:] The horizontal axis of humidity chart is dry-bulb temperature.

4. Wet-bulb temperature is based on the equilibrium between rates of energy transfer to the bulb and which of the following?
a) Evaporation of water
b) Saturation of water
c) Freezing of water
d) None of the mentioned

View Answer

Answer: a [Reason:] Wet-bulb temperature is based on the equilibrium between rates of energy transfer to the bulb and evaporation of water.

5. To plot the web-bulb line one information is wet-bulb temperature, what can be the other information?
a) Relative humidity
b) Dry-bulb temperature
c) Humidity
d) All of the mentioned

View Answer

Answer: d [Reason:] Relative humidity, Dry-bulb temperature, or Humidity any of these can be the other information.

6. Which of the following is also called adiabatic cooling?
a) Humidification
b) Humidity
c) Humid temperature
d) None of the mentioned

View Answer

Answer: a [Reason:] Humidification is also called adiabatic cooling.

7. What is the humidity at the point of intersection of dew point curve and wet-bulb line?
a) Dry-bulb Humidity
b) Wet-bulb Humidity
c) Specific Humidity
d) None of the mentioned

View Answer

Answer: a [Reason:] The humidity at the point of intersection of dew point curve and wet-bulb line is called the wet-bulb humidity.

8. What is the heat transferred from the system in humidification process?
a) Positive
b) Negative
c) Zero
d) None of the mentioned

View Answer

Answer: c [Reason:] Humidification is the adiabatic cooling process which means no heat exchange.

9. What happens to water in the humidification process?
a) Evaporates
b) Freezes
c) Saturated
d) None of the mentioned

View Answer

Answer: a [Reason:] Water evaporates during the humidification process.

10. What is the temperature of the point of intersection of adiabatic cooling line and specific humid volume line?
a) Wet-bulb temperature
b) Dry-bulb temperature
c) Dew point temperature
d) None of the mentioned

View Answer

Answer: b [Reason:] The temperature of the point of intersection of adiabatic cooling line and specific humid volume line is the dry-bulb temperature.

11. What is the enthalpy change of the mixture if the enthalpy change of air is 10 J, enthalpy change of water vapor is 15 J, and the humidity is 40%?
a) 10 J
b) 12 J
c) 14 J
d) 16 J

View Answer

Answer: d [Reason:] ∆H = ∆Hair + ∆Hwater(h) = 10 + 15*0.4 = 16 J.

12. What is the enthalpy change of 15oC system if the heat capacity of air at 20oC is 10 J/oC and heat capacity of water vapor at 20oC is 20 J/oC, enthalpy of vaporization is 20 J, and the humidity is 50%?
a) 50 J
b) 60 J
c) 80 J
d) 110 J

View Answer

Answer: d [Reason:] ∆H = 10(20 – 15) + 0.5*[20(20 – 15) + 20] = 50 + 60 = 110 J.

13. What is the enthalpy change of 10oC system if the heat capacity of air at 20oC is 5 J/oC and heat capacity of water vapor at 20oC is 15 J/oC, enthalpy of vaporization is 10 J, and the humidity is 40%?
a) 58 J
b) 84 J
c) 114 J
d) 135 J

View Answer

Answer: c [Reason:] ∆H = 5(20 – 10) + 0.4*[15(20 – 10) + 10] = 50 + 64 = 114 J.

14. What is the enthalpy change of 20oC system if the heat capacity of air at 25oC is 5 J/oC and heat capacity of water vapor at 25oC is 10 J/oC, enthalpy of vaporization is 10 J, and the humidity is 20%?
a) 25 J
b) 37 J
c) 44 J
d) 51 J

View Answer

Answer: b [Reason:] ∆H = 5(25 – 20) + 0.2*[10(25 – 20) + 10] = 25 + 12 = 37 J.

15. What is the enthalpy change of 30oC system if the heat capacity of air at 38oC is 5 J/oC and heat capacity of water vapor at 38oC is 10 J/oC, enthalpy of vaporization is 10 J, and the humidity is 60%?
a) 14 J
b) 39 J
c) 65 J
d) 94 J

View Answer

Answer: d [Reason:] ∆H = 5(38 – 30) + 0.6*[10(38 – 30) + 10] = 40 + 54 = 94 J.

Set 5

1. A bypass stream does not go through which of the following?
a) Mixer
b) Process
c) Separator
d) None of the mentioned

View Answer

Answer: b [Reason:] A bypass stream directly goes from divider to separator skipping the process.

2. Which of the following does the purge stream comes from?
a) Feed stream
b) Product stream
c) Recycle stream
d) None of the mentioned

View Answer

Answer: c [Reason:] Purge stream is the removal of unwanted material from the recycle stream.

3. Which of the following composition can be controlled by bypass stream?
a) Exit stream
b) Feed
c) Process
d) None of the mentioned

View Answer

Answer: a [Reason:] Bypass stream can be used to control the composition of final stream.

4. Which of the following composition is controlled by purge stream?
a) Feed
b) Product stream
c) Recycle stream
d) None of the mentioned

View Answer

Answer: c [Reason:] Composition of recycle stream can be controlled by purge stream.

5. Which of the following is decreased by the bypass stream?
a) Feed
b) Product
c) Feed & Product
d) None of the mentioned

View Answer

Answer: a [Reason:] Bypass stream decreases the feed and increases the product.

Answer question 6 – 10 for the following diagram.

chemical-engineering-questions-answers-bypass-purge-q6

6. What is the value of P?
a) 5 mole/hr
b) 10 mole/hr
c) 15 mole/hr
d) 20 mole/hr

View Answer

Answer: c [Reason:] Overall CH4 balance: 0.6*20 = 0.8*P, => P = 15 mole/hr.

7. What is the value of W?
a) 5 mole/hr
b) 10 mole/hr
c) 20 mole/hr
d) 25 mole/hr

View Answer

Answer: a [Reason:] Overall C2H6 balance: 0.4*20 = 0.2*15 + 1*W, => W = 5 mole/hr.

8. What is the value of G?
a) 12.5 mole/hr
b) 25 mole/hr
c) 37.5 mole/hr
d) 50 mole/hr

View Answer

Answer: a [Reason:] Material balance: G = 5 + H, CH4 balance: G(0.6) = H, solving both equations G = 12.5 mole/hr.

9. What is the value of H?
a) 5 mole/hr
b) 7.5 mole/hr
c) 10 mole/hr
d) 12.5 mole/hr

View Answer

Answer: [Reason:] H = 0.6G = 7.5 mole/hr.

10. What is the value of B?
a) 7.5 mole/hr
b) 15 mole/hr
c) 20 mole/hr
d) 22.5 mole/hr

View Answer

Answer: a [Reason:] B = F – G = 20 – 12.5 = 7.5 mole/hr.

Answer question 11 – 15 for the following diagram. The reaction is N2 + 3H2 -> 2NH3, overall conversion of N2 is 80%.

chemical-engineering-questions-answers-bypass-purge-q11
11. What is the value of W?
a) 2.5 mole
b) 5 mole
c) 9.5 mole
d) 20 mole

View Answer

Answer: a [Reason:] 0.2*10 = 0.8*1*W, => W = 2.5 mole.

12. What is the value of P?
a) 2.5 mole
b) 5 mole
c) 7.5 mole
d) 10 mole

View Answer

Answer: c [Reason:] F = W + P, => 10 = 2.5 + P, => P = 7.5 mole.

13. What is the fraction of N2 in purge stream?
a) 0.04
b) 0.05
c) 0.16
d) 0.4

View Answer

Answer: b [Reason:] N2 balance: 10*0.2 – 10*0.2*0.8 = 7.5*x, => x = 0.05.

14. How many moles of H2 are recycled?
a) 1.5
b) 2.2
c) 2.8
d) 3.9

View Answer

Answer: d [Reason:] Moles of H2 in G stream = 10*0.8 – 3*10*0.2*0.8 = 3.2 = moles of H2 in H stream. Moles of H2 in P stream = 0.95*7.5 = 7.125, => moles of H2 in R stream = 7.125 – 3.2 = 3.925.

15. What is the value of R?
a) 1.4
b) 2.8
c) 4.2
d) 6.8

View Answer

Answer: c [Reason:] Recycled moles of H2 = 3.925, recycled moles of N2 = (10*0.2 – 10*0.2*0.8) – 0.05 = 0.35, => R = 3.925 + 0.35 = 4.275.

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