Chemical Engineering Interview Questions and Answers for experienced 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
Answer: c
Clarification: 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
Answer: c
Clarification: 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
Answer: b
Clarification: 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
Answer: d
Clarification: 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
Answer: c
Clarification: 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 an amount of water evaporated and amount of input?
a) 2:1
b) 3:2
c) 5:4
d) 6:5
Answer: d
Clarification: 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%
Answer: b
Clarification: 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
Answer: c
Clarification: 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
Answer: b
Clarification: 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
Answer: b
Clarification: 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%
Answer: b
Clarification: 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