# Chemical Engineering Questions and Answers – Solving Material Balance Problems for Single Units

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 an aqueous solution with sulfur 1 g/L, what is the amount of sulfur in the organic compound after extraction?
a) 2 g/L
b) 5 g/L
c) 9 g/L
d) 15 g/L

Explanation: Let the amount of sulfur in the 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

Explanation: Let the rate of an 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

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

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

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.
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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 an amount of water evaporated and amount of input?
a) 2:1
b) 3:2
c) 5:4
d) 6:5

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%

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

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 an 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

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 the 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

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%

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

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 %

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%

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

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.

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