This set of Introduction to Chemical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Multiple Component Systems”.

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

View answer

Explanation: 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

View answer

Explanation: 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

View answer

Explanation: 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

View answer

Explanation: 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

View answer

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.

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

View answer

Explanation: 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%

View answer

Explanation: 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

View answer

Explanation: 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

View answer

Explanation: 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

View answer

Explanation: 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

View answer

Explanation: 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

View answer

Explanation: 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

View answer

Explanation: 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

View answer

Explanation: 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

View answer

Explanation: Percentage of O2 = 20/(10 + 15)*100 = 8%.

**Sanfoundry Global Education & Learning Series – Introduction to Chemical Engineering.**

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