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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1901. |
What is the change potential energy of a stone of mass 5 Kg that falls from a cliff 10 m high?(a) 5 J(b) 50 J(c) 500 J(d) 5000 J |
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Answer» The correct option is (c) 500 J Explanation: PE = 5*10*10 = 500 J. |
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| 1902. |
If the pressure of a gas at 27^oC is 15 mm Hg, what will be its pressure at 127^oC, use B = 5 X 10^3, C = 0, T is in kelvin and P is in mm Hg?(a) 579 mm Hg(b) 775 mm Hg(c) 961 mm Hg(d) 999 mm Hg |
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Answer» Right option is (c) 961 mm Hg The best explanation: ln P1/P2 = B (1/T2 – 1/T1), => ln 15/P2 = 5 X 10^3 (1/400 – 1/300), P2 = 961 mm Hg. |
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| 1903. |
What is the heat of dissolution if the heat of reaction is 10 J in the energy balance?(a) -10 J(b) 10 J(c) 0(d) Cannot be determined |
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Answer» The correct answer is (a) -10 J Easy explanation: Heat of dissolution = – Heat of solution = – Heat of reaction = -10 J. |
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| 1904. |
In a reactor two stream are entering, one with H2 at 10 atm and 27^oC, and another with O2 at 10 atm and 0^oC, producing H2O2, H2O, and O2 at 10 atm and 23^oC, what are the total degrees of freedom?(a) 12(b) 17(c) 19(d) 23 |
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Answer» Correct answer is (b) 17 The explanation: Total species = 1 + 1 + 3 = 5, total stream flows = 3, stream pressures = 3, stream temperatures = 3, heat released = 1, and extent of reaction = 2, => degrees of freedom = 5 + 3 + 3 + 3 + 1 + 2 = 17. |
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| 1905. |
What is the heat transferred from the system if the output sensible heat is 5 J, input sensible heat is 25 J, and heat of solution is 20 J?(a) 0(b) 5 J(c) 10 J(d) 15 J |
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Answer» Correct choice is (a) 0 For explanation I would say: Q = ∆Hout,sensible – ∆Hin,sensible + ∆Hsolution = 5 – 15 + 20 = 15 J. |
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| 1906. |
Pressure of a gas at temperature T is P, What is the pressure of the gas at 2T, use B = 1?(a) Pe^1/T(b) Pe^1/2T(c) Pe^1/3T(d) Pe^1/4T |
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Answer» Right answer is (b) Pe^1/2T Explanation: ln P/P2 = 1*(1/2T – 1/T), => P2 = Pe^1/2T. |
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| 1907. |
What is the heat transferred from the system if the output sensible heat is 10 J, input sensible heat is 15 J, and heat of solution is 20 J?(a) 10 J(b) 15 J(c) 20 J(d) 25 J |
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Answer» Right answer is (b) 15 J Easiest explanation: Q = ∆Hout,sensible – ∆Hin,sensible + ∆Hsolution = 10 – 15 + 20 = 15 J. |
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| 1908. |
What is the value of P in the following process?(a) 5 mole(b) 15 mole(c) 25 mole(d) 35 mole |
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Answer» The correct choice is (c) 25 mole Explanation: 10 + 20 = 5 + P, => P = 25 mole. |
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| 1909. |
What is the sensible heat of a reaction if the enthalpy of outputs is 5 J, and enthalpy of inputs is 10 J?(a) -5 J(b) 5 J(c) -10 J(d) 10 J |
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Answer» Right option is (a) -5 J To explain I would say: Q = ∆Houtput – ∆Hinput = 5 – 10 = -5J. |
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| 1910. |
A gas mixture has 10% CH4 with pressure 5 Pa at 27^oC at the rate F, is passed through a reactor with another stream containing 60% CH4 with pressure 5 Pa at -73^oC at the rate 10 liter/s, if the percentage of CH4 with pressure 5 Pa at 27^oC in product is 80%, what is the value of F?(a) 1 liter/s(b) 1.4 liter/s(c) 2 liter/s(d) 2.6 liter/s |
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Answer» The correct choice is (b) 1.4 liter/s Explanation: Overall material balance: F + 10 = P, Overall CH4 Balance: 0.1*F*5/300 + 0.6*10*5/200 = 0.8*P*5/300, => F + 90 = 8P, => F = 10/7 = 1.4 liter/s. |
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| 1911. |
In a reactor two stream are entering, one with H2 at 10 atm and 27^oC, and another with O2 at 10 atm and 0^oC, producing H2O at 10 atm and 23^oC, what are the total degrees of freedom?(a) 10(b) 14(c) 16(d) 19 |
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Answer» The correct answer is (b) 14 Easy explanation: Total species = 1 + 1 + 1 = 3, total stream flows = 3, stream pressures = 3, stream temperatures = 3, heat released = 1, and extent of reaction = 1, => degrees of freedom = 3 + 3 + 3 + 3 + 1 + 1 = 14. |
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| 1912. |
The feed of a process simulator is 2 mole with enthalpy 4 J/mole, the heat input is 6 J, and the work performed is 10 J, what will be the product enthalpy if a mole of the product are 2?(a) 1 J/mole(b) 2 J/mole(c) 3 J/mole(d) 5 J/mole |
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Answer» Right choice is (b) 2 J/mole The best I can explain: Energy balance, 2*H = 2*4 + 6 – 10 = 4, => H = 2 J/mole. |
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| 1913. |
The feed of a process simulator is 5 mole with enthalpy 3 J/mole, the heat input is 15 J, and the work performed is 20 J, what will be the product enthalpy if a mole of the product are 2?(a) 1 J/mole(b) 2 J/mole(c) 3 J/mole(d) 5 J/mole |
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Answer» The correct answer is (d) 5 J/mole To elaborate: Energy balance, 2*H = 5*3 + 15 – 20 = 10, => H = 5 J/mole. |
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| 1914. |
What is the heat transferred from 2 mole of the system if the output sensible heat is 3 J/mole, input sensible heat is 2 J/mole, and heat of solution is 2 J/mole?(a) 2 J(b) 3 J(c) 5 J(d) 6 J |
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Answer» The correct answer is (d) 6 J The best I can explain: Q = ∆Hout,sensible – ∆Hin,sensible + ∆Hsolution = 2*(3 – 2 + 2) = 6 J. |
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| 1915. |
What is the heat transferred from 5 mole of the system if the output sensible heat is 1 J/mole, input sensible heat is 3 J/mole, and heat of solution is 4 J/mole?(a) 0(b) 5 J(c) 10 J(d) 20 J |
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Answer» Correct answer is (c) 10 J Easiest explanation: Q = ∆Hout,sensible – ∆Hin,sensible + ∆Hsolution = 5*(1 – 3 + 4) = 10 J. |
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| 1916. |
The feed of a process simulator is 10 mole with enthalpy 2 J/mole, the heat input is 15 J, and the work performed is 10 J, what will be the product enthalpy if mole of the product are 5?(a) 2 J/mole(b) 3 J/mole(c) 4 J/mole(d) 5 J/mole |
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Answer» Correct option is (d) 5 J/mole For explanation: Energy balance, 5*H = 10*2 + 15 – 10 = 25, => H = 5 J/mole. |
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| 1917. |
CO with pressure 10 atm at 27^oC is passed through a reactor with O2, what is the pressure of CO2 produced if 60% CO reacted?(a) 2 atm(b) 6 atm(c) 10 atm(d) 14 atm |
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Answer» The correct option is (b) 6 atm To explain I would say: The reaction is 2CO + O2 -> 2CO2, pressure of CO2 = 10*0.6 = 6 atm. |
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| 1918. |
In a random reaction 2 mole of CH4 and 1 mole of C2H4 are reacted, and 1 mole of C3H8 and 1 mole of CxHy are produced, what is x*y?(a) 2(b) 4(c) 8(d) 12 |
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Answer» The correct choice is (b) 4 The best explanation: Material balance for C: 2*1 + 1*2 =1*3 +1*x, => x = 1, for H: 2*4 + 1*4 = 1*8 + 1*y, => y = 4, => x*y = 4. |
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| 1919. |
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 |
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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. |
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| 1920. |
A reactor is supplied with 10 mole of Hexane and some hydrogen to produce 20% methane, 30% propane and 50% butane, what is the rate of products?(a) 10 mole(b) 16.6 mole(c) 26.6 mole(d) 40 mole |
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Answer» Correct answer is (c) 26.6 mole Easiest explanation: The reaction is 2C6H14 + 2H2 -> CH4 + 2C3H8 + C5H12, Let the rate of products be P. Element balance, C: 2*10(6) = 0.2P(1) + 2*0.3P(3) + 0.5P(5), => 120 = 4.5P, => P = 26.6 mole. |
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| 1921. |
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 |
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Answer» The correct choice is (b) Consumption + Output For explanation I would say: Accumulation = Input – Output + Generation – Consumption, => x = Consumption + Output. |
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| 1922. |
What is the value of P in the following process?(a) 5 mole(b) 10 mole(c) 15 mole(d) 20 mole |
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Answer» Right answer is (c) 15 mole The best I can explain: 10 + 20 = P + 10 + 5, => P = 15 mole. |
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| 1923. |
Pentane is burnt with 100% of excess air, what is the percentage of H2O in the products?(a) 12.4%(b) 34.6%(c) 42.1%(d) 56.9% |
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Answer» Right option is (c) 42.1% To explain: The reaction is C5H12 + 8O2 -> 5CO2 + 6H2O. Basis: 10 moles of C5H12, => moles of O2 reacted = 80, => moles of O2 entered the process = 160, => moles of CO2, H2O and O2 in products are 50, 60 and 80, => percentage of H2O = 80/190*100 = 42.1%. |
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| 1924. |
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 |
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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. |
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| 1925. |
A reactor is supplied with glucose and oxygen to produce 40% CO2, 60% H2O, what is the ratio of glucose reacted and products?(a) 0.2(b) 0.4(c) 0.6(d) 0.8 |
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Answer» Correct option is (b) 0.4 Best explanation: The reaction is C6H12O6 + 6O2 -> 6CO2 + 6H2O, Let the rate of glucose and products be F and P, Element balances, C: F(6) = 6*0.4P(1), => F/P = 0.4. |
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| 1926. |
A reactor is supplied with Hexane and hydrogen to produce 20% methane, 30% propane and 50% butane, what is the ratio of hydrogen consumed and the hexane reacted?(a) 0.21(b) 0.45(c) 0.59(d) 0.73 |
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Answer» Right option is (d) 0.73 To explain I would say: The reaction is 2C6H14 + 2H2 -> CH4 + 2C3H8 + C5H12, Let the rate of Hexane, hydrogen and products be F, W and P respectively. Element balances, C: 2F(6) = 0.2P(1) + 2*0.3P(3) + 0.5P(5), => 8F = 3P, H: 2F(14) + 2W(2) = 0.2P(4) + 2*0.3P(8) + 0.5P(12), => 7F + W = 2.9P, => W/F= 2.2/3 = 0.73. |
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| 1927. |
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) 100 |
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Answer» The correct choice is (c) 75 To explain: Percentage of O2 = 15/(10 + 10)*100 = 75%. |
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| 1928. |
A reactor supplied with CO2 and H2O, the product contains CO2, H2O and H2CO3, if the rate of products is 10 mole and percentage of CO2 in products is 20%, what is the feed rate of H2O?(a) 2 mole(b) 4 mole(c) 8 mole(d) 10 mole |
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Answer» Right choice is (c) 8 mole Easy explanation: The reaction is CO2 + H2O -> H2CO3. Let the rate of H2O be W, and percentage of H2O in products be x. Material balances, H: W(2) = x*10(2) + (1 – x – 0.2)*10(2), => W = 8 mole. |
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| 1929. |
A reactor is supplied with 10 moles glucose and excess oxygen to produce CO2 and H2O, if the rate of products is 25 moles, what is the percentage of CO2 in the products?(a) 0.1(b) 0.2(c) 0.3(d) 0.4 |
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Answer» The correct choice is (d) 0.4 Easy explanation: The reaction is C6H12O6 + 6O2 -> 6CO2 + 6H2O. Let the fraction of CO2 be x, Element balances, C: 10(6) = 6*x*5(1), => x = 0.4. |
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| 1930. |
Methane is at the rate 50 moles is supplied to a reactor with air at the rate 1000 moles, the reactor leaves O2, N2, CO2 and H2O, what is the percentage of N2 in products?(a) 25.4%(b) 45.2%(c) 75.2%(d) 95.6% |
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Answer» The correct option is (c) 75.2% Easy explanation: The reaction is CH4 + 2O2 -> CO2 + 2H2O. O2 supplied = 1000*21/100 = 210 moles, N2 supplied = 1000*79/100 = 790 moles, => O2 unreacted = 210 – 100 = 110 moles, CO2 formed = 50 moles, H2O formed = 100 moles. => Percentage of N2 = 790/1050*100 = 75.2%. |
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| 1931. |
Methane is at the rate 50 moles is supplied to a reactor with air at the rate 1000 moles, the reactor leaves O2, N2, CO2 and H2O, what is the percentage of O2 in products?(a) 10.4%(b) 25.8%(c) 45.6%(d) 78.2% |
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Answer» Right choice is (a) 10.4% Easiest explanation: The reaction is CH4 + 2O2 -> CO2 + 2H2O. O2 supplied = 1000*21/100 = 210 moles, N2 supplied = 1000*79/100 = 790 moles, => O2 unreacted = 210 – 100 = 110 moles, CO2 formed = 50 moles, H2O formed = 100 moles. => Percentage of O2 = 110/1050*100 = 10.4%. |
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| 1932. |
Methane is at the rate 50 moles is supplied to a reactor with air at the rate 1000 moles, the reactor leaves O2, N2, CO2 and H2O, what is the percentage of H2O in products?(a) 5.8%(b) 9.5%(c) 15.4%(d) 20.7% |
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Answer» Right choice is (b) 9.5% The explanation is: The reaction is CH4 + 2O2 -> CO2 + 2H2O. O2 supplied = 1000*21/100 = 210 moles, N2 supplied = 1000*79/100 = 790 moles, => Osub>2 unreacted = 210 – 100 = 110 moles, COsub>2 formed = 50 moles, Hsub>2O formed = 100 moles. => Percentage of H2O = 100/1050*100 = 9.5%. |
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| 1933. |
Methane is at the rate 50 moles is supplied to a reactor with air at the rate 1000 moles, the reactor leaves O2, N2, CO2 and H2O, what is the percentage of CO2 in products?(a) 4.7%(b) 10.4%(c) 16.5%(d) 24.3% |
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Answer» Right choice is (a) 4.7% The explanation is: The reaction is CH4 + 2O2 -> CO2 + 2H2O. O2 supplied = 1000*21/100 = 210 moles, N2 supplied = 1000*79/100 = 790 moles, => O2 unreacted = 210 – 100 = 110 moles, CO2 formed = 50 moles, H2O formed = 100 moles. => Percentage of CO2 = 50/1050*100 = 4.7%. |
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| 1934. |
Methane is at the rate 50 moles is supplied to a reactor with air at the rate 1000 moles, the reactor leaves O2, N2, CO2 and H2O, what is the rate of products?(a) 900 moles(b) 950 moles(c) 1000 moles(d) 1050 moles |
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Answer» Right choice is (d) 1050 moles For explanation: The reaction is CH4 + 2O2 -> CO2 + 2H2O. O2 supplied = 1000*21/100 = 210 moles, N2 supplied = 1000*79/100 = 790 moles, => O2 unreacted = 210 – 100 = 110 moles, CO2 formed = 50 moles, H2O formed = 100 moles. => Rate of products = 790 + 110 + 50 + 100 = 1050 moles. |
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| 1935. |
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) 62.5 |
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Answer» Correct answer is (d) 62.5 To explain I would say: Percentage of O2 = 10/(10 + 6)*100 = 62.5%. |
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| 1936. |
A container has gas mixture of 40% CO2 and 60% H2 with total pressure 10 Pa, if now O2 is added to it isothermally so that the final composition of the mixture becomes 20% CO2, 30% H2, and 50% O2, what is the partial pressure of O2 in the new composition?(a) 5 Pa(b) 10 Pa(c) 20 Pa(d) 35 Pa |
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Answer» The correct choice is (b) 10 Pa For explanation: Partial pressure of H2 = 0.6*10 = 6 Pa, => Partial pressure of O2 = 0.5*6/0.3 = 10 Pa. |
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| 1937. |
A reactor supplied with CO2 and H2O, the product contains H2O and H2CO3, if the rate of H2O is 10 mole, what is the rate of products?(a) 5 mole(b) 10 mole(c) 15 mole(d) 20 mole |
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Answer» The correct option is (b) 10 mole For explanation I would say: The reaction is CO2 + H2O -> H2CO3. Let the rate of products be P, and the fraction of H2CO3 in products be x, material balances, H: 10(2) = x*P(2) + (1 – x)*P(2), => P = 10 mole. |
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| 1938. |
A reactor supplied with CO2 and H2O, the product contains H2O and H2CO3, if the rate of CO2 is 10 mole, what is the percentage of H2CO3 in products?(a) 10(b) 50(c) 75(d) 100 |
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Answer» The correct answer is (d) 100 To elaborate: The reaction is CO2 + H2O -> H2CO3. Let the rate of CO2 and H2O and product be F and W respectively, and the fraction of H2CO3 in products be x. Element balances, C: F(1) = x*10(1), => F = 10x, => x = 1, => percentage of H2CO3 = 100%. |
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| 1939. |
A reactor is supplied with 10 moles glucose and excess oxygen to produce 40% CO2, 60% H2O, what is the rate of products?(a) 25 mole(b) 50 mole(c) 75 mole(d) 100 mole |
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Answer» Correct answer is (a) 25 mole Explanation: The reaction is C6H12O6 + 6O2 -> 6CO2 + 6H2O, Let the rate of products be P. Element balances, C: 10(6) = 6*0.4P(1), => P = 25 mole. |
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| 1940. |
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) 0.96 |
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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 = (1*0.6 + 9*0.2)/(1 + 9) = 0.24. |
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| 1941. |
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) 50 |
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Answer» Right choice is (d) 50 The explanation is: Percentage of O2 = 4/(4 + 4)*100 = 50%. |
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| 1942. |
Octane is burnt with 40% excess O2, what is the percentage of CO2 in products?(a) 18.18%(b) 36.36%(c) 54.54%(d) 72.72% |
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Answer» Correct choice is (b) 36.36% The best I can explain: The reaction is C8H18 + 12.5O2 -> 8CO2 + 9H2O. Basis: 10 moles of C8H18, => moles of O2 reacted = 125, => moles of O2 entered the process = 175, => moles of CO2, H2O and O2 in products are 80, 90 and 50, => percentage of CO2 = 80/220*100 = 36.36%. |
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| 1943. |
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/hr |
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Answer» Correct option is (d) 2 Kg/hr The best I can explain: Rate at product side = 10*50/100 = 5 Kg/hr, => rate of CO2 = 5*0.4 = 2 Kg/hr. |
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| 1944. |
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/hr |
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Answer» Correct choice is (c) 25 Kg/hr To elaborate: Total feed rate = Total product rate, => 10 + 30 = 15 + x, => x = 25 Kg/hr. |
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| 1945. |
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/hr |
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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. |
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| 1946. |
A reactor supplied with CO2 and H2O, the product contains 40% CO2, 30% H2O and 30% H2CO3, what is the ratio of feed rate of CO2 and rate of products?(a) 0.3(b) 0.4(c) 0.7(d) 0.9 |
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Answer» Right answer is (c) 0.7 Explanation: The reaction is CO2 + H2O -> H2CO3. Let the rate of CO2, H2O and H2CO3 be F, W and P respectively. Element balances, C: F(1) = 0.4P(1) + 0.3P(1), => F/P = 0.7. |
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| 1947. |
Propane is burnt with 20% excess O2, what is the percentage of CO2 in products?(a) 12.5%(b) 25%(c) 45%(d) 75.5% |
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Answer» Correct option is (a) 12.5% Explanation: The reaction is C3H8 + 5O2 -> 3CO2 + 4H2O. Basis: 10 moles of C3H8, => moles of O2 reacted = 50, => moles of O2 entered the process = 60, => moles of CO2, H2O and O2 in products are 30, 40 and 10, => percentage of CO2 = 10/80*100 = 12.5%. |
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| 1948. |
CH4 is burnt with 100% excess O2 at atmospheric pressure, if CH4 conversion is 100%, what is the pressure at dew point?(a) 0.2 atm(b) 0.4 atm(c) 0.6 atm(d) 0.8 atm |
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Answer» Correct choice is (b) 0.4 atm The best I can explain: CH4 + 2O2 -> CO2 + 2H2O, Basis: 100 moles of CH4, => moles of O2 reacted = 200, => Product contains, moles of CO2 = 100, moles of H2O = 200, moles of O2 = 200. => Fraction of H2O = 0.4, => Pressure at due point = 1*0.4 = 0.4 atm. |
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| 1949. |
N2 with pressure 10 atm and H2 with pressure 30 atm, both with volume 10 liter at 27^oC are passed into a reactor, how many moles NH3 is produced?(a) 4(b) 8(c) 12(d) 16 |
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Answer» Correct choice is (b) 8 For explanation: The reaction is N2 + 3H2 -> 2NH3, moles of N2 = 10*10/(0.0821*300) = 4, moles of H2 = 30*10/(0.0821*300) = 12, => extent of reaction = (0 – 4)/(-1) = 4, => moles of NH3 = 4*2 = 8. |
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| 1950. |
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% |
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Answer» Correct answer is (c) 60% The explanation is: Efficiency = 9/(10 + 5)*100 = 60%. |
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