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

1. Feed of a reactor has 0.25 mass fraction of H_{2}O and 0.75 mass fraction of CO_{2} with rate 4 g/hr, if the product rate is 20 g/hr what is the mass of CO_{2} in the product side?

a) 3 grams

b) 9 grams

c) 15 grams

d) 21 grams

View answer

Explanation: Mass fraction of CO

_{2}will be same on both sides, => mass of CO

_{2}in the product side = 20*0.75 = 15 grams.

2. A reactor has feed of two streams, stream 1 has CO_{2} and H_{2}O with mass fraction of 0.4 and 0.6 and rate 10 kg/hr, and stream 2 has CO_{2} and H_{2}O with mass fraction 0.1 and 0.9 and rate 40 Kg/hr, what the mass fraction of CO_{2} in the product stream?

a) 0.08

b) 0.16

c) 0.24

d) 0.32

View answer

Explanation: Mass fraction of CO

_{2}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 O_{2} and some SO_{2}, if the product rate is 1000 Kg/hr then what is the rate of SO_{2} 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 SO

_{2}= (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 O_{2} and O_{3} 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 O_{2} 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 CO_{2} and SO_{2} with mass fraction 0.4 and 0.6 with the rate 10 Kg/hr, what is the rate of CO_{2} 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 CO

_{2}= 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 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/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% CH_{4} and 80% H_{2}O is 5 mole/hr, the contained was initially filled with 3 mole of H_{2}O, what will be the percentage of CH_{4} in product stream?

a) 5

b) 12.5

c) 25

d) 37.5

View answer

Explanation: Percentage of CH

_{4}= 1/(1 + 4 + 3)*100 = 12.5%.

12. The rate of flow in stream with 20% O_{2} and 80% CO_{2} is 5 mole/hr, the contained was initially filled with 3 mole of O_{2}, what will be the percentage of O_{2} in product stream?

a) 12.5

b) 25

c) 37.5

d) 50

View answer

Explanation: Percentage of O

_{2}= 4/(4 + 4)*100 = 50%.

13. The rate of flow in stream with 40% O_{2} and 60% CO_{2} is 10 mole/hr, the contained was initially filled with 6 mole of O_{2}, what will be the percentage of O_{2} in product stream?

a) 12.5

b) 25

c) 50

d) 62.5

View answer

Explanation: Percentage of O

_{2}= 10/(10 + 6)*100 = 62.5%.

14. The rate of flow in stream with 50% O_{2} and 50% CO_{2} is 10 mole/hr, the contained was initially filled with 10 mole of O_{2}, what will be the percentage of O_{2} in product stream?

a) 25

b) 50

c) 75

d) 100

View answer

Explanation: Percentage of O

_{2}= 15/(10 + 10)*100 = 75%.

15. The rate of flow in stream with 75% O_{2} and 25% CO_{2} is 20 mole/hr, the contained was initially filled with 5 mole of O_{2}, what will be the percentage of O_{2} in product stream?

a) 20

b) 40

c) 60

d) 80

View answer

Explanation: Percentage of O

_{2}= 20/(10 + 15)*100 = 8%.

**Sanfoundry Global Education & Learning Series – Basic Chemical Engineering.**

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