90 + Interview Questions in Introduction To Material Balances in Basic Chemical Engineering Page 1 InterviewSolution

1.	In a batch process, 5 mole of SO2 and 3 mole of CO2 entered a vessel with some SO2, if 10 mole of product leaves the vessel, what is the percentage of SO2 in the product?(a) 10%(b) 30%(c) 50%(d) 70%I have been asked this question by my college director while I was bunking the class.I need to ask this question from Material Balances for Batch and Semi-Batch Processes topic in section Introduction to Material Balances of Basic Chemical Engineering
Answer» The CORRECT ANSWER is (d) 70% For EXPLANATION I would say: Percentage of H2 in the product = (10 – 3)/10*100 = 70%.

Discussion

2.	In a batch process, 2 mole of SO2 and 3 mole of CO2 entered a vessel with some SO2, if 5 mole of product leaves the vessel, what is the percentage of SO2 in the product?(a) 10%(b) 20%(c) 40%(d) 50%I had been asked this question in semester exam.My question is based upon Material Balances for Batch and Semi-Batch Processes topic in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» The CORRECT choice is (c) 40% For EXPLANATION I WOULD say: Percentage of H2 in the product = 2/5*100 = 40%.

Discussion

3.	In a batch process, 5 mole of H2 and 3 mole of CO2 entered a vessel with some CO2, if 10 mole of product leaves the vessel, what is the percentage of H2 in the product?(a) 10%(b) 30%(c) 50%(d) 80%I had been asked this question by my college professor while I was bunking the class.I'm obligated to ask this question of Material Balances for Batch and Semi-Batch Processes topic in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct choice is (c) 50% The EXPLANATION is: PERCENTAGE of H2 in the product = 5/10*100 = 50%.

Discussion

4.	In a batch process, 40 mole of O2 and 30 mole of CO2 entered a vessel with some CO2, if 100 mole of product leaves the vessel, what is the percentage of O2 in the product?(a) 10%(b) 40%(c) 60%(d) 80%The question was asked in exam.I'd like to ask this question from Material Balances for Batch and Semi-Batch Processes topic in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» RIGHT OPTION is (b) 40% Explanation: Percentage of H2O in the PRODUCT = 40/100*100 = 40%.

Discussion

5.	In a batch process, 50 Kg of H2O and 40 Kg of CO2 entered a vessel with some H2O, if 100 Kg of product leaves the vessel, what is the percentage of H2O in the product?(a) 10%(b) 30%(c) 60%(d) 80%I had been asked this question by my college director while I was bunking the class.My question is based upon Material Balances for Batch and Semi-Batch Processes topic in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» The CORRECT OPTION is (c) 60% Explanation: PERCENTAGE of H2O in the product = (100 – 40)/100*100 = 60%.

Discussion

6.	10 Kg of oil is burnt in a container and taken off from the container if 1% of oil remains on the container wall, how much oil is taken off?(a) 8.9 Kg(b) 9 Kg(c) 9.9 Kg(d) 10 KgThis question was posed to me in class test.I would like to ask this question from Material Balances for Batch and Semi-Batch Processes in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» CORRECT choice is (c) 9.9 KG For EXPLANATION: Amount of oil TAKEN off = 10*(1 – 1/100) = 9.9 Kg.

Discussion

7.	Two reactants enter a vessel and a reaction is carried out, it is an example of which of the following?(a) Batch(b) Semi-batch(c) Neither of them(d) Batch & Semi-batchI got this question in class test.I would like to ask this question from Material Balances for Batch and Semi-Batch Processes in section Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct option is (B) Semi-batch To ELABORATE: SINCE material ENTERS the vessel and operated but does not comes out, so it is a semi-batch process.

Discussion

8.	In a batch process, 70 Kg of H2O and 10 Kg of NaOH and some other reactants enters the vessel, if 100 Kg of the product leaves the vessel, what is the percentage of H2O in the product?(a) 10%(b) 30%(c) 70%(d) 100%The question was posed to me during an online exam.This question is from Material Balances for Batch and Semi-Batch Processes in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» The correct CHOICE is (C) 70% The BEST explanation: Percentage of H2O in the product = (70/100)*100 = 70%.

Discussion

9.	Water boiling in a container is an example of which of the following?(a) Batch(b) Semi-batch(c) Neither of them(d) Batch & Semi-batchI got this question by my college director while I was bunking the class.I want to ask this question from Material Balances for Batch and Semi-Batch Processes in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» RIGHT choice is (c) Neither of them Explanation: Since NOTHING is entering in the container but STEAM is leaving the container, it is neither a batch PROCESS nor a semi-batch process.

Discussion

10.	Water flowing in plants from tap through a pipe, the process is an example of which of the following process?(a) Batch(b) Semi-batch(c) Neither of them(d) Batch & Semi-batchI have been asked this question during an online exam.This question is from Material Balances for Batch and Semi-Batch Processes in section Introduction to Material Balances of Basic Chemical Engineering
Answer» Right choice is (a) Batch Easiest explanation: SINCE water enters the pipe from one END and leaves from ANOTHER end, it is a batch PROCESS.

Discussion

11.	A semi-batch process differs from a closed process in which of the following?(a) Input(b) Output(c) Generation(d) ConsumptionThe question was posed to me during an online interview.This key question is from Material Balances for Batch and Semi-Batch Processes in division Introduction to Material Balances of Basic Chemical Engineering
Answer» RIGHT OPTION is (a) INPUT Explanation: A semi-batch PROCESS has some input, while a CLOSED process does not have input.

Discussion

12.	Tap filling a jar with water is an example of which of the following process?(a) Batch(b) Semi-batch(c) Neither of them(d) Batch & Semi-batchThis question was addressed to me in an interview for internship.The question is from Material Balances for Batch and Semi-Batch Processes in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» The CORRECT option is (b) Semi-batch The BEST I can EXPLAIN: Since water is entering the jar and does not leave, it is a semi-batch PROCESS.

Discussion

13.	A semi-batch process differs from an open process in which of the following?(a) Input(b) Output(c) Generation(d) ConsumptionI had been asked this question in homework.Question is taken from Material Balances for Batch and Semi-Batch Processes in division Introduction to Material Balances of Basic Chemical Engineering
Answer» Right choice is (B) OUTPUT For explanation: A semi-batch PROCESS does not have output, while an open SYSTEM has output.

Discussion

14.	A semi-batch process is that in which material enters the system and _____ which of the following should fill the gap?(a) Leaves(b) Does not leaves(c) Half of it leaves(d) None of the mentionedThe question was posed to me by my college director while I was bunking the class.My enquiry is from Material Balances for Batch and Semi-Batch Processes topic in section Introduction to Material Balances of Basic Chemical Engineering
Answer» Right OPTION is (b) Does not leaves For EXPLANATION: A semi-batch PROCESS is that in which MATERIAL enters a system and does not leave.

Discussion

15.	A batch process is that in which material enters the system and _____ which of the following should fill the gap?(a) Leaves(b) Does not leaves(c) Half of it leaves(d) None of the mentionedI have been asked this question in class test.Asked question is from Material Balances for Batch and Semi-Batch Processes topic in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» Right CHOICE is (a) Leaves The BEST explanation: A BATCH process is that in which material ENTERS the system and operated and leaves.

Discussion

16.	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/hrThis question was addressed to me in an online interview.My query is from Accounting for Chemical Reactions in Material Balances topic in division Introduction to Material Balances of Basic Chemical Engineering
Answer» The correct choice is (a) 1 mole/hr Explanation: The reaction is C6H6 + 3H2 -> C6H12, => flow out RATE of C6H12 = 3*1/3 = 1 mole/hr.

Discussion

17.	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/hrThis question was posed to me during an online exam.This interesting question is from Accounting for Chemical Reactions in Material Balances in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct OPTION is (b) 1.2 mole/hr To explain I would say: The REACTION is C3H8 + 5O2 -> 3CO2 + 4H2O, => flow out rate of CO2 = 2*3/5 = 1.2 mole/hr.

Discussion

18.	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/hrI had been asked this question in examination.Enquiry is from Accounting for Chemical Reactions in Material Balances topic in division Introduction to Material Balances of Basic Chemical Engineering
Answer» The CORRECT choice is (b) 2 mole/hr Best explanation: The reaction is H2SO4 + 2NaCl -> 2HCL + Na2SO4, => FLOW out rate of HCl = 2*1 = 2 mole/hr.

Discussion

19.	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/hrThis question was addressed to me in class test.My question comes from Accounting for Chemical Reactions in Material Balances in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» The CORRECT choice is (a) 1 mole/hr To explain: The reaction is CH4 + 2O2 -> CO2 + 2H2O, => FLOW out rate of CO2 = 1*1 = 1 mole/hr.

Discussion

20.	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/hrThis question was posed to me in an internship interview.I'd like to ask this question from Accounting for Chemical Reactions in Material Balances topic in division Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct OPTION is (a) 1 mole/hr For explanation: The reaction is HCl + NaOH -> NACL + H2O, => FLOW out rate of NaCl = 1 mole/hr.

Discussion

21.	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 molarI have been asked this question during an online interview.I'm obligated to ask this question of Accounting for Chemical Reactions in Material Balances in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» The correct answer is (b) 5 molar Explanation: The BALANCED chemical reaction of the system is CH4 + O2 -> CO2 + 2H2O, => AMOUNT of CO2 = 5 molar.

Discussion

22.	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 molarThe question was asked at a job interview.I would like to ask this question from Accounting for Chemical Reactions in Material Balances topic in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» Right OPTION is (a) 2.5 molar Explanation: The BALANCED chemical reaction of the SYSTEM is 2KNO3 + H2SO4 -> K2SO4 + 2HNO3, => amount of K2SO4 = 2.5 molar.

Discussion

23.	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 molarThis question was addressed to me in an internship interview.Asked question is from Accounting for Chemical Reactions in Material Balances in division Introduction to Material Balances of Basic Chemical Engineering
Answer» The correct choice is (a) 1 MOLAR For explanation I WOULD say: The balanced reaction of the system is 2NaCl + H2CO3 -> Na2CO3 + 2HCl, => amount of Na2CO3 = 1 molar.

Discussion

24.	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/minThe question was asked in an interview for job.Asked question is from Accounting for Chemical Reactions in Material Balances in division Introduction to Material Balances of Basic Chemical Engineering
Answer» Right answer is (c) 300 L/min Explanation: As the system is in STEADY state, => rate of PRODUCT = rate of feed, => rate of NA2CO3 = 200 + 100 = 300 L/min.

Discussion

25.	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 mentionedThe question was posed to me in final exam.Enquiry is from Accounting for Chemical Reactions in Material Balances in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct ANSWER is (c) Non-reactive Easiest explanation: As the FORMULA does not INVOLVE GENERATION and consumption it is a non-reactive SYSTEM.

Discussion

26.	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 molarThe question was asked during an online exam.Question is from Accounting for Chemical Reactions in Material Balances in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» The CORRECT answer is (a) 1 molar To explain I would say: The balanced reaction of the system is NAOH + HCL -> NaCl + H2O, => amount of NaCl = amount of HCl = amount of NaOH = 1 molar.

Discussion

27.	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/minThis question was addressed to me during a job interview.Query is from Accounting for Chemical Reactions in Material Balances topic in section Introduction to Material Balances of Basic Chemical Engineering
Answer» The CORRECT OPTION is (B) 200 mole/min Explanation: As the system is in steady-state, rate of product = rate of feed, => rate of NACL = 100 + 100 = 200 mole/min.

Discussion

28.	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 mentionedI had been asked this question in an internship interview.The origin of the question is Accounting for Chemical Reactions in Material Balances in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» The CORRECT OPTION is (b) Closed The BEST I can EXPLAIN: As the FORMULA does not involve input and output it is a closed system.

Discussion

29.	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 mentionedI have been asked this question at a job interview.Asked question is from Accounting for Chemical Reactions in Material Balances topic in division Introduction to Material Balances of Basic Chemical Engineering
Answer» Right choice is (d) None of the mentioned To explain I WOULD say: 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.

Discussion

30.	The rate of flow in stream with 50% O2 and 50% CO2 is 10 mole/hr, the contained was initially filled with 10 mole of O2, what will be the percentage of O2 in product stream?(a) 25(b) 50(c) 75(d) 100The question was posed to me in an online quiz.My doubt is from Multiple Component Systems topic in division Introduction to Material Balances of Basic Chemical Engineering
Answer» The CORRECT choice is (c) 75 To explain: PERCENTAGE of O2 = 15/(10 + 10)*100 = 75%.

Discussion

31.	The rate of flow in stream with 75% O2 and 25% CO2 is 20 mole/hr, the contained was initially filled with 5 mole of O2, what will be the percentage of O2 in product stream?(a) 20(b) 40(c) 60(d) 80This question was posed to me in my homework.I'd like to ask this question from Multiple Component Systems topic in section Introduction to Material Balances of Basic Chemical Engineering
Answer» The correct answer is (d) 80 Easy explanation: Percentage of O2 = 20/(10 + 15)*100 = 8%.

Discussion

32.	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 mentionedThe question was posed to me in class test.My doubt is from Accounting for Chemical Reactions in Material Balances in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» The correct choice is (B) Consumption + Output For explanation I would say: ACCUMULATION = INPUT – Output + Generation – Consumption, => X = Consumption + Output.

Discussion

33.	The rate of flow in stream with 40% O2 and 60% CO2 is 10 mole/hr, the contained was initially filled with 6 mole of O2, what will be the percentage of O2 in product stream?(a) 12.5(b) 25(c) 50(d) 62.5I had been asked this question in class test.The doubt is from Multiple Component Systems in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» CORRECT ANSWER is (d) 62.5 To EXPLAIN I would say: PERCENTAGE of O2 = 10/(10 + 6)*100 = 62.5%.

Discussion

34.	The rate of flow in stream with 20% O2 and 80% CO2 is 5 mole/hr, the contained was initially filled with 3 mole of O2, what will be the percentage of O2 in product stream?(a) 12.5(b) 25(c) 37.5(d) 50I had been asked this question by my college director while I was bunking the class.This is a very interesting question from Multiple Component Systems topic in section Introduction to Material Balances of Basic Chemical Engineering
Answer» Right CHOICE is (d) 50 The EXPLANATION is: Percentage of O2 = 4/(4 + 4)*100 = 50%.

Discussion

35.	The rate of flow in stream with 20% CH4 and 80% H2O is 5 mole/hr, the contained was initially filled with 3 mole of H2O, what will be the percentage of CH4 in product stream?(a) 5(b) 12.5(c) 25(d) 37.5I had been asked this question during an online interview.Enquiry is from Multiple Component Systems topic in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» The correct answer is (b) 12.5 The best explanation: Percentage of CH4 = 1/(1 + 4 + 3)*100 = 12.5%.

Discussion

36.	Water flowing with TSS at the rate of 40 Liter/min, is discharged into a river flowing at the rate of 20 Liter/min, if the concentration of TSS in the river is 5 mg/L, what is the concentration of TSS in water?(a) 5 mg/L(b) 10 mg/L(c) 15 mg/L(d) 20 mg/LThis question was posed to me in quiz.I need to ask this question from Multiple Component Systems in section Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct answer is (B) 10 mg/L The explanation is: CONCENTRATION of TSS in water = 5*40/20 = 10 mg/L.

Discussion

37.	Water flowing with chlorine concentration 10 mg/L at the rate of 20 Liter/min, is discharged into a river if the concentration of chlorine in a river is 5 mg/L what is the flow rate of river?(a) 10 Liter/min(b) 20 Liter/min(c) 30 Liter/min(d) 40 Liter/minI got this question during a job interview.I would like to ask this question from Multiple Component Systems topic in section Introduction to Material Balances of Basic Chemical Engineering
Answer» The correct OPTION is (a) 10 Liter/min Explanation: Rate of FLOW of river = 20*5/10 = 10 Liter/min.

Discussion

38.	Waste water flowing with TSS concentration 40 mg/L at the rate of 10 Liter/min is discharged into the river if the river is flowing at the rate of 15 Liter/min, what is the concentration of TSS in the river?(a) 10 mg/L(b) 30 mg/L(c) 45 mg/L(d) 60 mg/LThis question was addressed to me in exam.I want to ask this question from Multiple Component Systems topic in section Introduction to Material Balances of Basic Chemical Engineering
Answer» Right OPTION is (d) 60 mg/L The BEST I can explain: Concentration of TSS in RIVER = 40*15/10 = 60 mg/L.

Discussion

39.	A reactor with efficiency 50%, has the feed CO2 and SO2 with mass fraction 0.4 and 0.6 with the rate 10 Kg/hr, what is the rate of CO2 in product side?(a) 0.5 Kg/hr(b) 1 Kg/hr(c) 1.5 Kg/hr(d) 2 Kg/hrI got this question in a national level competition.My doubt stems from Multiple Component Systems in section Introduction to Material Balances of Basic Chemical Engineering
Answer» CORRECT OPTION is (d) 2 Kg/hr The best I can explain: RATE at product side = 1050/100 = 5 Kg/hr, => rate of CO2 = 50.4 = 2 Kg/hr.

Discussion

40.	A reactor has feed rate of 10 Kg/ hr and 5 Kg/hr and product rate of 9 Kg/hr, what is the efficiency of the reactor?(a) 10%(b) 30%(c) 60%(d) 100%I got this question in a job interview.This question is from Multiple Component Systems in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» CORRECT answer is (c) 60% The EXPLANATION is: EFFICIENCY = 9/(10 + 5)*100 = 60%.

Discussion

41.	A reactor has two feed streams with the ratio of their rate as 1:9 and has O2 and O3 with the mass fraction of 0.6 and 0.4 in stream 1 and 0.2 and 0.8 in stream 2, what is the mass fraction of O2 in the product side?(a) 0.24(b) 0.48(c) 0.72(d) 0.96This question was posed to me in exam.My doubt is from Multiple Component Systems in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» The correct answer is (a) 0.24 The explanation: LET the STREAM rates be 1 Kg/hr and 9 Kg/hr, => mass FRACTION of O¬2 in product side = (10.6 + 90.2)/(1 + 9) = 0.24.

Discussion

42.	A reactor has feed from 2 streams, one with rate 10 Kg/hr and other with 30 Kg/hr, and product side also has 2 streams one of which is at the rate of 15 Kg/hr, what is the rate of other product stream?(a) 5 Kg/hr(b) 15 Kg/hr(c) 25 Kg/hr(d) 35 Kg/hrI had been asked this question in an interview.My question is taken from Multiple Component Systems topic in chapter Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct choice is (C) 25 Kg/hr To elaborate: TOTAL FEED rate = Total PRODUCT rate, => 10 + 30 = 15 + x, => x = 25 Kg/hr.

Discussion

43.	A reactor has feed of two streams, stream 1 has CO2 and H2O with mass fraction of 0.4 and 0.6 and rate 10 kg/hr, and stream 2 has CO2 and H2O with mass fraction 0.1 and 0.9 and rate 40 Kg/hr, what the mass fraction of CO2 in the product stream?(a) 0.08(b) 0.16(c) 0.24(d) 0.32I have been asked this question in semester exam.My enquiry is from Multiple Component Systems topic in section Introduction to Material Balances of Basic Chemical Engineering
Answer» Right ANSWER is (b) 0.16 For explanation: MASS fraction of CO2 in product STREAM = (0.410 + 0.140)/(10 + 40) = 0.16.

Discussion

44.	Feed of a reactor has 0.75 mass fraction of O2 and some SO2, if the product rate is 1000 Kg/hr then what is the rate of SO2 in the feed?(a) 100 Kg/hr(b) 250 Kg/hr(c) 750 Kg/hr(d) 1000 Kg/hrThis question was posed to me during an online exam.The origin of the question is Multiple Component Systems topic in division Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct option is (B) 250 Kg/hr The BEST explanation: Feed rate = Product rate = 1000 Kg/hr, => Rate of SO2 = (1 – 0.75)*1000 = 250 Kg/hr.

Discussion

45.	Feed of a reactor has 0.25 mass fraction of H2O and 0.75 mass fraction of CO2 with rate 4 g/hr, if the product rate is 20 g/hr what is the mass of CO2 in the product side?(a) 3 grams(b) 9 grams(c) 15 grams(d) 21 gramsI got this question during an interview.I'd like to ask this question from Multiple Component Systems in division Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct answer is (c) 15 grams The best explanation: Mass FRACTION of CO2 will be same on both SIDES, => mass of CO2 in the product SIDE = 20*0.75 = 15 grams.

Discussion

46.	An unsteady state system, the flow in the rate of A is 6 mole/s, what is the flow out rate of B if the accumulation was 18 mole in 9 seconds?(a) 1 mole/s(b) 2 mole/s(c) 3 mole/s(d) 4 mole/sThis question was posed to me in homework.Enquiry is from Steady-State and Unsteady-State Systems topic in division Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct CHOICE is (d) 4 mole/s Best explanation: FLOW rate of B = 6 – 18/9 = 4 mole/s.

Discussion

47.	An unsteady state system, the flow in rate of A is 5 mole/s, what is the flow out rate of B if the accumulation was 10 mole in 5 seconds?(a) 1 mole/s(b) 2 mole/s(c) 3 mole/s(d) 5 mole/sThis question was posed to me in a job interview.I want to ask this question from Steady-State and Unsteady-State Systems topic in section Introduction to Material Balances of Basic Chemical Engineering
Answer» The CORRECT choice is (c) 3 mole/s The explanation: FLOW RATE of B = 5 – 10/5 = 3 mole/s.

Discussion

48.	An unsteady state system, the flow in the rate of A is 12 mole/s, what is the flow out rate of B if the accumulation was 18 mole in 3 seconds?(a) 4 mole/s(b) 6 mole/s(c) 8 mole/s(d) 9 mole/sThe question was asked by my college professor while I was bunking the class.My doubt stems from Steady-State and Unsteady-State Systems topic in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct CHOICE is (B) 6 mole/s To explain: Flow RATE of B = 12 – 18/3 = 6 mole/s.

Discussion

49.	An unsteady state system, the flow in rate of A is 10 mole/s, what is the flow out rate of B if the accumulation was 10 mole in 2 seconds?(a) 5 mole/s(b) 10 mole/s(c) 15 mole/s(d) 20 mole/sI have been asked this question during an interview for a job.Query is from Steady-State and Unsteady-State Systems topic in section Introduction to Material Balances of Basic Chemical Engineering
Answer» Correct ANSWER is (a) 5 mole/s Explanation: Flow RATE of B = 10 – 10/2 = 5 mole/s.

Discussion

50.	An unsteady state system, the flow in rate of A is 8 mole/s, what is the flow out rate of B if the accumulation was 6 mole in 2 seconds?(a) 1 mole/s(b) 2 mole/s(c) 3 mole/s(d) 5 mole/sThe question was asked in final exam.The origin of the question is Steady-State and Unsteady-State Systems in portion Introduction to Material Balances of Basic Chemical Engineering
Answer» RIGHT ANSWER is (d) 5 mole/s The explanation: Flow rate of B = 8 – 6/2 = 5 mole/s.

Discussion

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