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# Multiple choice question for engineering

## Set 1

1. Mechanical energy satisfy which of the following?
a) Completely Convertible
b) Non-convertible
c) Partially convertible
d) None of the mentioned

Answer: a [Reason:] Mechanical energy is completely convertible by an ideal engine.

2. Which of the following is not an example of mechanical energy?
a) Kinetic energy
b) Potential energy
c) Internal energy
d) Work

Answer: c [Reason:] Internal energy is not a mechanical energy.

3. Which of the following allows complete conversion of mechanical energy?
a) Frictionless process
b) Dissipative process
c) Mixing of components
d) All of the mentioned

Answer: a [Reason:] A frictionless process is reversible so it allows complete conversion of mechanical energy.

4. Which process allows complete conversion of mechanical energy?
a) Irreversible
b) Reversible
c) None of the mentioned
d) Both of the mentioned

Answer: b [Reason:] Reversible process so it allows complete conversion of mechanical energy.

5. The mechanical energy balance is the product of equation of motion and which of the following?
a) Mass
b) Velocity
c) Volume
d) None of the mentioned

Answer: b [Reason:] The mechanical energy balance is the product of equation of motion and velocity.

6. Bernoulli equation is valid for which process?
a) Reversible
b) Irreversible
c) None of them
d) Both of them

Answer: a [Reason:] Bernoulli equation is valid for the process in which loss of mechanical energy is zero, which is reversible.

7. What is the height difference of the water (density = 1000 Kg/m3), if the change in velocity is 2 m/s and change in pressure is 100 Pa?
a) 0.21 m
b) 0.42 m
c) 0.56 m
d) 0.84 m

Answer: a [Reason:] ∆h = (100/1000 + 22/2)/10 = 0.21 m.

8. What is the height difference of the water (density = 1000 Kg/m3), if the change in velocity is 10 m/s and change in pressure is 500 Pa?
a) 1.01 m
b) 5.05 m
c) 10.1 m
d) 20.2 m

Answer: b [Reason:] ∆h = (500/1000 + 102/2)/10 = 5.05 m.

9. What is the height difference of the water (density = 1000 Kg/m3), if the change in velocity is 10 m/s and change in pressure is 1000 Pa?
a) 1.4 m
b) 2.9 m
c) 4.2 m
d) 5.1 m

Answer: d [Reason:] ∆h = (1000/1000 + 102/2)/10 = 5.05 m.

10. What is the height difference of the water (density = 1000 Kg/m3), if the change in velocity is 6 m/s and change in pressure is 2000 Pa?
a) 1.2
b) 2.0
c) 3.6
d) 4.4

Answer: b [Reason:] ∆h = (2000/1000 + 62/2)/10 = 2.0 m.

11. What is the reversible work done if 5 m3 of an ideal gas is compressed isothermally from 1 Pa to 5 Pa?
a) 4.41 J
b) 8.01 J
c) 10.20 J
d) 14.18 J

Answer: b [Reason:] V2 = 1*5/5 = 1 m3, => W = –51pdv = –51 (5/v) dv = 5 ln5 = 8.04 J.

12. What is the reversible work done if 4 m3 of an ideal gas is compressed isothermally from 2 Pa to 8 Pa?
a) 2.01 J
b) 6.11 J
c) 8.56 J
d) 11.09 J

Answer: d [Reason:] V2 = 2*4/8 = 1 m3, => W = –41pdv = –41 (8/v) dv = 8 ln4 = 11.09 J.

13. What is the reversible work done if 8 m3 of an ideal gas is compressed isothermally from 4 Pa to 16 Pa?
a) 12.48 J
b) 28.55 J
c) 44.36 J
d) 56.96 J

Answer: c [Reason:] V2 = 8*4/16 = 2 m3, => W = –82pdv = –82 (32/v) dv = 32 ln4 = 44.36 J.

14. What is the reversible work done if 4 m3 of an ideal gas is compressed adiabatically pV1.5 = K from 2 Pa to 16 Pa?
a) 4 J
b) 16 J
c) 24 J
d) 56 J

Answer: b [Reason:] V2 = (2*41.5/16)1/1.5 = 1 m3, => W = –41pdv = –41 (16/v1.5) dv = 16 J.

15. What is the reversible work done if 4 m3 of an ideal gas is compressed adiabatically pV2.5 = K from 1 Pa to 32 Pa?
a) 18.6 J
b) 20.4 J
c) 24.6 J
d) 26.2 J

Answer: a [Reason:] V2 = (1*42.5/32)1/2.5 = 1 m3, => W = –41pdv = –41 (32/v2.5) dv = 18.6 J.

## Set 2

1. What is the pressure of a gas at T = 227oC, it is given that A = 5, B = 4 X 104 and C = 0 if T is in kelvin and p in mm Hg?
a) 0.024 mm Hg
b) 0.049 mm Hg
c) 0.075 mm Hg
d) 0.096 mm Hg

Answer: b [Reason:] ln p* = A – B/(C + T), => ln p = 5 – 4 X 103/(273 +227) = -3, => p = 0.049 mm Hg.

2. A material’s mass is given by the equation, m = 32P0.02/T0.5, where T is in Kelvin, m is in Kg and P is in mm Hg, what is the mass of material at 127oC, use A = 10, B = 6 X 103 and C = 0?
a) 1.44 Kg
b) 1.69 Kg
c) 1.96 Kg
d) 2.25 Kg

Answer: a [Reason:] ln p* = A – B/(C + T) = 10 – 6 X 103/(400) = -5, => p = 0.0067 mm Hg, => m = 32*(0.0067)0.02/4000.5 = 1.44 Kg.

3. If the pressure of a gas at 27oC is 15 mm Hg, what will be its pressure at 127oC, use B = 5 X 103, 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

Answer: c [Reason:] ln P1/P2 = B (1/T2 – 1/T1), => ln 15/P2 = 5 X 103 (1/400 – 1/300), P2 = 961 mm Hg.

4. Pressure of a gas at 127oC is 10 mm Hg and the pressure at 527oC is 20 Pa, what is the value of B?
a) 225.1
b) 365.5
c) 499.2
d) 554.5

Answer: d [Reason:] ln 10/20 = B (1/800 – 1/400), => B = 554.5.

5. Pressure of a gas at temperature T is P, What is the pressure of the gas at 2T, use B = 1?
a) Pe1/T
b) Pe1/2T
c) Pe1/3T
d) Pe1/4T

Answer: b [Reason:] ln P/P2 = 1*(1/2T – 1/T), => P2 = Pe1/2T.

6. What is the pressure of CO2 at 325 K, if it is given that pressure at 330 K is 6 Pa, and pressure at 320 K is 4 Pa?
a) 4.5 Pa
b) 5 Pa
c) 5.5 Pa
d) 6 Pa

Answer: b [Reason:] Using linear interpolation, P325K = 4 + (325 – 320)*(6 – 4)/(330 – 320) = 4 + 5*2/10 = 5 Pa.

7. Temperature of SO2 at 5 Pa is 395 K and at 10 Pa temperature is 420 K, what is the temperature at 6 Pa?
a) 400 K
b) 405 K
c) 410 K
d) 415 K

Answer: a [Reason:] Using linear interpolation, T6Pa = 395 + (6 – 5)*(420 – 395)/(10 – 5) = 395 + 5 = 400 K.

8. At which of the following values of pressure a solid melts first?
a) 5 atm
b) 720 mm Hg
c) 500 KPa
d) 40 Torr

Answer: a [Reason:] A solid melts first at the highest pressure which is 5 atm.

9. A liquid evaporates first at which of the following conditions?
a) High attraction forces
b) Low attraction forces
c) Low vapor pressure
d) High boiling point

Answer: b [Reason:] A liquid evaporates first at low attraction forces.

10. At triple point, vapor pressure of ice is given by ln p* = 18 – 2560/T, and vapor pressure of water is given by ln p* = 28 – 5120/T, what is the temperature at triple point?
a) 225 K
b) 256 K
c) 320 K
d) 400 K

Answer: b [Reason:] At triple point pressure and temperature of all the phases becomes same, => 18 – 2560/T = 28 – 5120/T, => 10 = 2560/T, => T = 256 K.

11. The vapor pressure of liquid benzene is given by ln p* = 15 – 1280/T, and that of vapor benzene is given by ln p* = 23 – 2560/T, what is the boiling point of benzene?
a) 160 K
b) 200 K
c) 240 K
d) 300 K

Answer: a [Reason:] At boiling point pressure and temperature of both the phases will be same, => 23 – 2560/T = 15 – 1280/T, => 8 = 1280/T, => T = 160K.

12. The vapor pressure of a compound in liquid phase is given by ln p* = 21 – 2800/T, and in solid phase it is given by ln p* = 28 – 3500/T, what is the sublimation temperature of the compound?
a) 100 K
b) 120 K
c) 150 K
d) 200 K

Answer: a [Reason:] At sublimation point pressure and temperature of both the phases will be same, => 21 – 2800/T = 28 – 3500/T, => T = 100 K.

13. For CaOCl2, the rate of evaporation is given by W = 325p0.5M/T1.5 g/liter, where T is in K, p in mm Hg, and M is molecular weight, if its pressure is given by ln p* = 25 – 2700/T, what is the evaporation rate at T = 150 K?
a) 256 g/liter
b) 499 g/liter
c) 571 g/liter
d) 744 g/liter

Answer: d [Reason:] ln p* = 25 – 2700/150 = 7, => p = 1096.6 mm Hg, => W = 325(1096.6)0.5*127/1501.5 = 744 g/liter.

14. What is the vapor pressure of ammonia at 305 K, if it is given that the vapor pressure of ammonia at 315 K is 5 mm Hg and at 325 K it is 6 mm Hg?
a) 1 mm Hg
b) 2 mm Hg
c) 3 mm Hg
d) 4 mm Hg

Answer: d [Reason:] p305K = 5 – (315 – 305)*(6 – 5)/(325 – 315) = 5 – 1 = 4 mm Hg.

15. The pressure of a compound at 200 K is 10 mm Hg and at 400 K it is 50 mm Hg, what is the pressure at T = 250 K?
a) 18.1 mm Hg
b) 24.2 mm Hg
c) 36.3 mm Hg
d) 48.4 mm Hg

Answer: a [Reason:] Let the pressure of the compound is given by ln p* = a + b/T, => ln 10 = a + b/200, ln 50 = a + b/400, => ln 5 = -b/400, => b = -643.7, => a = 5.5, => At T = 250 K, ln p* = 5.5 – 643.7/250 = 2.9, p = 18.1 mm Hg.

## Set 3

1. What is the mole fraction of sodium with moles n in a solution with moles of solvent N?
a) N/n
b) n/N
c) n/(n + N)
d) N/(n + N)

Answer: c [Reason:] Mole fraction is the ratio of moles of solvent and total number of moles of solution, => mole fraction = n/(n + N).

2. What is the unit of mole fraction?
a) Kg/m3
b) N/m3
c) m-2
d) None of the mentioned

Answer: d [Reason:] Since the mole fraction is ratio of the moles of solute and the moles of solution, it has no unit.

3. If the mole fraction of 4.5 moles of solute is 0.5, what is the number of moles of solvent?
a) 9
b) 4.5
c) 1.5
d) 3

Answer: b [Reason:] 0.5 = 4.5/(4.5 + n), => n = 4.5.

4. What is the mass fraction of 4 moles of Sodium in 8 moles of Hydrogen chloride?
a) 0.24
b) 0.48
c) 0.56
d) 0.72

Answer: a [Reason:] Mass of sodium = 4*23 = 92 g, mass of hydrogen chloride = 8*36.5 = 292 g, => mass fraction of sodium = 92/(92 + 292) = 0.24.

5. If the mole fraction of sodium is 0.25 in 9 moles of liquid carbon dioxide, what is the mass fraction of sodium?
a) 0.10
b) 0.15
c) 0.20
d) 0.25

Answer: b [Reason:] 0.25 = n/(n + 9), => n = 3, => mass of sodium = 3*23/(3*23 + 9*44) = 0.15.

6. A solution has 25% w/w solution of Sodium Chloride in 10 grams of water what is the mole fraction of Sodium Chloride?
a) 0.05
b) 0.06
c) 0.07
d) 0.08

Answer: c [Reason:] Mass of sodium chloride = 10*25/100 = 2.5 grams, => moles of sodium chloride = 2.5/58.5 = 0.0427, moles of water = 10/18 = 0.555, => mole fraction = 0.0427/(0.0427 + 0.555) = 0.07.

7. If the mole fraction of SO2 in H2SO4 is 0.2, what is the ratio of mass of sulfur and oxygen in the solution?
a) 0.166
b) 6
c) 1.8
d) 0.555

Answer: d [Reason:] Let moles of SO2 and H2SO4 be x and y respectively, => x/(x + y) = 0.2, => y/x = 4, => ratio of mass of sulfur and oxygen in solution = (x*32 + y*32)/ (x*2(16) + y*4(16)) = 0.555.

8. A mixture of gases SO2, O2 and 4 g of N2 and total mass of mixture is 10 g, what is the sum of mole fraction of SO2 and O2?
a) 0.25
b) 0.5
c) 0.75
d) 1

Answer: c [Reason:] Sum of the mole fraction of all gases = 1, => sum of mole fraction of SO2 and O2 = 1 – 4/10 = 0.75.

9. A neutral gas that is made of nitrogen and hydrogen only contains 7 g of nitrogen is in a mixture of with total mass 17 g, what is the mass fraction of the gas?
a) 0.5
b) 0.25
c) 0.45
d) 0.8

Answer: a [Reason:] The neutral gas that is made of nitrogen and hydrogen is NH3, => moles of nitrogen = moles of NH3 = 7/14 = 0.5, => mass of hydrogen = 1.5*1 = 1.5 g => total mass of NH3 = 7 + 1.5 = 8.5 g, => mass fraction = 8.5/17 = 0.5.

10. A mixture contains 23 g of ethanol and 36 g of water, some acetic acid is added to the mixture and its mole fraction becomes 0.5, how much acetic acid was added?
a) 25 g
b) 75 g
c) 100 g
d) 150 g

Answer: d [Reason:] Let m grams of acetic acid was added, => 0.5 = (m/60)/{(23/46) + (36/18) + (m/60)}, => m = 150 g.

11. A mixture contains 9 gram each of H2O and NaCl, what is the mole fraction of NaCl?
a) 0.15
b) 0.23
c) 0.45
d) 0.64

Answer: b [Reason:] Mole fraction of NaCl = (9/58.5)/[9/58.5 + 9/18] = 0.23.

12. A mixture contains 10 grams each of CH4, H2O, and SO2, what is the mole fraction of CH4?
a) 0.14
b) 0.31
c) 0.46
d) 0.67

Answer: c [Reason:] Mole fraction of CH4 = (10/16)/[(10/16) + (10/18) + (10/64)] = 0.46.

13. A mixture contains 80% of HCl and 20% H2O by weight, what is the mole fraction of HCl?
a) 0.15
b) 0.29
c) 0.45
d) 0.66

Answer: d [Reason:] Mole fraction of HCl = (80/36.5)/[(80/36.5) + (20/18)] = 0.66.

14. A mixture contains 80% of NaOH and 20% H2O by weight, what is the mole fraction of NaOH?
a) 0.24
b) 0.42
c) 0.64
d) 0.78

Answer: c [Reason:] Mole fraction of HCl = (80/40)[(80/40) + (20/18)] = 0.64.

15. A compound contains 40% of Sulphur and 60% of Oxygen, what is the empirical formula of the compound?
a) SO2
b) SO3
c) SO4
d) None of the mentioned

Answer: b [Reason:] Mole fraction of Sulphur = (40/32)[(40/32) + (60/16)] = 0.25, => Mole fraction of oxygen = 0.75, => Empirical formula of compound is SO3.

## Set 4

1. 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 grams

Answer: c [Reason:] Mass fraction of CO2 will be same on both sides, => mass of CO2 in the product side = 20*0.75 = 15 grams.

2. 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) 0.32

Answer: b [Reason:] Mass fraction of CO2 in product stream = (0.4*10 + 0.1*40)/(10 + 40) = 0.16.

3. 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

Answer: c [Reason:] Total feed rate = Total product rate, => 10 + 30 = 15 + x, => x = 25 Kg/hr.

4. 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

Answer: b [Reason:] Feed rate = Product rate = 1000 Kg/hr, => Rate of SO2 = (1 – 0.75)*1000 = 250 Kg/hr.

5. 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

Answer: a [Reason:] 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.

6. 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

Answer: d [Reason:] Rate at product side = 10*50/100 = 5 Kg/hr, => rate of CO2 = 5*0.4 = 2 Kg/hr.

7. 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%

Answer: c [Reason:] Efficiency = 9/(10 + 5)*100 = 60%.

8. 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/L

Answer: d [Reason:] Concentration of TSS in river = 40*15/10 = 60 mg/L.

9. 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 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/min

Answer: a [Reason:] Rate of flow of river = 20*5/10 = 10 Liter/min.

10. 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/L

Answer: b [Reason:] Concentration of TSS in water = 5*40/20 = 10 mg/L.

11. 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) 37.5

Answer: b [Reason:] Percentage of CH4 = 1/(1 + 4 + 3)*100 = 12.5%.

12. 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

Answer: d [Reason:] Percentage of O2 = 4/(4 + 4)*100 = 50%.

13. 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

Answer: d [Reason:] Percentage of O2 = 10/(10 + 6)*100 = 62.5%.

14. 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

Answer: c [Reason:] Percentage of O2 = 15/(10 + 10)*100 = 75%.

15. 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) 80

Answer: d [Reason:] Percentage of O2 = 20/(10 + 15)*100 = 8%.

## Set 5

1. Which of the following is constant in a closed system?
a) Energy
b) Mass
c) Temperature
d) Momentum

Answer: b [Reason:] A material neither leaves nor enters a closed system.

2. How is the boundary wall of a closed system?
a) Impermeable
b) Permeable
c) Rigid
d) None of the mentioned

Answer: a [Reason:] Since the matter is not allowed to cross the boundary wall of a closed system, so the wall is impermeable.

3. Desert Cooler is an example of which of the following?
a) Closed System
b) Open System
c) Isolated System
d) None of the mentioned

Answer: b [Reason:] Since water is filled and then it vaporizes in desert cooler, this means material is flowing in and out of the system, so it is an open system.

4. A jar filled with water is an example of which of the following?
a) Closed System
b) Open System
c) Isolated System
d) None of the mentioned

Answer: a [Reason:] Since water is neither leaving nor entering the jar, this means it is a closed system.

5. A pressure cooker is an example of which of the following?
a) Closed System
b) Open System
c) Isolated System
d) None of the mentioned

Answer: b [Reason:] Since steam is expelled outside in the cooker as the temperature increase, it is an open system.

6. A country is an example of which of the following?
a) Closed System
b) Open System
c) Isolated System
d) None of the mentioned

Answer: b [Reason:] Since a man can leave or enter a country it is an example of open system.

7. An egg is an example of which of the following?
a) Closed System
b) Open System
c) Isolated System
d) None of the mentioned

Answer: a [Reason:] Since matter is neither entering nor leaving an egg, it’s a closed system.

8. The human body is an example of which of the following?
a) Closed System
b) Open System
c) Isolated System
d) None of the mentioned

Answer: b [Reason:] Since air, water and food is going in and out of a human body, it’s an open system.

9. What is the accumulation in the system after 10 second if the out flow and in flow rates are 10 g/s and 15 g/s?
a) 5 g
b) 10 g
c) 50 g
d) 100 g

Answer: c [Reason:] Accumulation = 10*(15 – 10) = 50 g.

10. What is the accumulation in the system after 4 second if the out flow and in flow rates are 3 g/s and 8 g/s?
a) 12 g
b) 20 g
c) 32 g
d) 40 g

Answer: b [Reason:] Accumulation = 4*(8 – 3) = 20 g.

11. The flow in rate of A is 5 mole/s, it undergoes a reaction A -> 2B, what are the moles of B that comes out after 5 seconds?
a) 5
b) 10
c) 25
d) 50

Answer: d [Reason:] Moles of B = 5*5*2 = 50.

12. The flow in rate of A is 8 mole/s, it undergoes a reaction 3A -> B, what are the moles of B that comes out after 6 seconds?
a) 5
b) 8
c) 10
d) 16

Answer: d [Reason:] Moles of B = 8*6/3 = 16.

13. The flow in rate of A and B are 5 mole/s and 10 mole/s respectively, which under goes reaction A + 2B -> 3C, what is the flow out rate of C?
a) 5 mole/s
b) 10 mole/s
c) 15 mole/s
d) 20 mole/s

Answer: c [Reason:] Rate of C = 3*5 = 15 mole/s.

14. The flow in rate of A and B are 4 mole/s and 8 mole/s respectively, which under goes reaction A + B -> 3C, what is the flow out rate of C?
a) 4 mole/s
b) 8 mole/s
c) 12 mole/s
d) 24 mole/s