This set of Chemical Engineering Interview Questions and Answers for experienced focuses on “Solving Material Balance Problems for Single Units”.

1. An aqueous solution with sulfur 10 g/L at the rate is 100 L/min and an organic compound with no sulfur at the rate 50 L/min were put into an extraction machine and produced aqueous solution with sulfur 1 g/L, what is the amount of sulfur in organic compound after extraction?

a) 2 g/L

b) 5 g/L

c) 9 g/L

d) 15 g/L

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Explanation: Let the amount of sulfur in organic compound be x. Sulfur balance equation, 100(10) + 50(0) = 100(1) + 50(x), => x = 9 g/L.

2. An aqueous solution with chlorine 30 g/L and an organic compound with no chlorine at the rate 50 L/min were put into an extraction machine and produced aqueous solution with chlorine 5 g/L and organic compound with chlorine 10 g/L, what is the rate of aqueous solution?

a) 10 L/min

b) 20 L/min

c) 30 L/min

d) 40 L/min

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Explanation: Let the rate of aqueous solution be x. Chlorine balance equation, x(30) + 50(0) = x(5) + 50(10), => x = 20 L/min.

3. A gas mixture input is given to a membrane with 40% O2 and 60% N2, the waste contains 80% of the input and the product contains 25% O2 and 75% N2, what is the percentage of O2 in the waste?

a) 0.22

b) 0.44

c) 0.66

d) 0.88

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Explanation: Let the input is 100 Kg mol, => waste = 0.8(100) = 80 Kg mol, => Product = 20 Kg mol. Equation of material balance, O2: 0.4(100) = 0.25(20) + y(80), => y = 0.4375 ≈ 0.44.

4. A gas mixture input is given to a membrane with 20% O2 and 80% N2, the waste contains 60% of the input and the product contains 30% O2 and 70% N2, what are the number of moles of O2 in the waste?

a) 1

b) 2

c) 3

d) Can’t be determined

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Explanation: Since the amount of input is not given, number of moles of O2 cannot be determined.

5. 100 Kg mol of a gas mixture input is given to a membrane with 30% O2 and 70% N2, the waste contains 60% of the input and the product contains 10% O2 and 90% N2, what are the number of moles of O2 in the waste?

a) 12

b) 16

c) 34

d) 42

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Explanation: Let the moles of O2 in the waste be x, Amount of waste = 100(0.6) = 60 Kg mol, => amount of product = 40 Kg mol. Material balance equation for O2: 0.3(100) + 0.1(40) = x, => x = 34 moles.

6. A liquid solid mixture with 10% solid and 90% liquid is input to a dryer, some of the liquid evaporated and the product contains 60% solid and 40% liquid, what is the ratio of amount of water evaporated and amount of input?

a) 2:1

b) 3:2

c) 5:4

d) 6:5

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Explanation: Let the amount of input be F and the amount of evaporated water be W. Solid balance: 0.1*F = 0*W + 0.6*(F – W), => F/W = 6/5.

7. A liquid solid mixture with 20% solid and 80% liquid is input to a dryer, if the amount of evaporated water is 60% of the amount of input, what is the percentage of solid in the product?

a) 40%

b) 50%

c) 80%

d) 90%

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Explanation: Let the amount of input be F, => the amount of evaporated water be 0.6F, => amount of product = 0.4F. Solid Balance: 0.2(F) = 0(0.6F) + x(0.4F), => x = 0.5, => percentage of solid in the product = 50%.

8. A 100 Kg liquid solid mixture with 10% solid and 90% liquid is input to a dryer, if the amount of evaporated water is 60% of the amount of input, what is the amount of liquid in the product?

a) 10 Kg

b) 20 Kg

c) 30 Kg

d) 40 Kg

View answer

Explanation: Let the amount of liquid in the product be x. Liquid balance: 0.9(100) = 1(60) + x, => x = 30 Kg.

9. 50 Kg of a solid liquid mixture containing 10% solid and 90% water is left open in atmosphere, after some time the water is 80%, what is the weight of the mixture now?

a) 10 Kg

b) 25 Kg

c) 35 Kg

d) 50 Kg

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Explanation: Let the new weight of mixture be P. Solid balance: 0.1(50) = 0.2P, => P = 25 Kg.

10. 50 Kg of a solid liquid mixture containing 20% solid and 80% water is left open in atmosphere, after some time the water is 60%, how much water is evaporated?

a) 15 Kg

b) 25 Kg

c) 35 Kg

d) 45 Kg

View answer

Explanation: Let the weight of the water evaporated be W. Water Balance: 0.8(50) = 1(W) + 0.6(50 – W), => W = 25 Kg.

11. A gas mixture of 40% He and 60% Ne is passed through a diffusion tube and the product has 10% He and 90% Ne, what is the percentage of He recovered?

a) 25%

b) 33.3%

c) 50%

d) 66.6%

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Explanation: Let the amount of input gas mixture be F, amount of He recovered be D. He balance: 0.4F = 1D + 0.1(F – D), => D/F = 0.333, => percentage of He recovered = 33.3%.

12. 100 Kg of a gas mixture of 40% He and 60% Ne is passed through a diffusion tube and the product has 20% He and 80% Ne, what is the amount of He recovered?

a) 25 Kg

b) 45 Kg

c) 50 Kg

d) 75 Kg

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Explanation: Let the amount of He recovered be D, He Balance: 0.4(100) = 1(D) + 0.2(100 – D), => D = 25 Kg.

13. A gas mixture of 40% He and 60% Ne is passed through a diffusion tube, if the amount of He is recovered is 20% of the input, what is the percentage of He in the product?

a) 25 %

b) 33.3 %

c) 50 %

d) 66.6 %

View answer

Explanation: Let the amount of input be F, He balance: 0.4(F) = 1(0.2F) + x(F – 0.2F), => x = 0.25, => Percentage of He in the product = 25%.

14. 400 g of CaSO4 is dissolved with 500 g of H2O, if 226 g of CaSO4.5H2O crystallizes out, then what is the percentage of CaSO4 in the remaining solution?

a) 39.16%

b) 51.54%

c) 62.28%

d) 75.67%

View answer

Explanation: Moles of crystal = 226/226 = 1, => the crystal has 136 g of CaSO4 and 90 g of water, => Remaining solution has 264 g of CaSO4 and 410 g of H2O, => Percentage of CaSO4 in remaining solution = 264/(264 + 410) *100 = 39.16%.

15. How much nitrogen must be added to a 20% nitrogen solution to obtain 100 Kg of 40% nitrogen solution?

a) 10 Kg

b) 25 Kg

c) 50 Kg

d) 75 Kg

View answer

Explanation: Let the amount of nitrogen added be x, and amount of 20% nitrogen solution be P. Nitrogen balance: x + 0.2(P) = 0.4(100), Water balance: 0.8(P) = 0.6(100), => P = 75 Kg, => x = 0.4(100) – 0.2(75), => x = 25 Kg.

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

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