250+ TOP MCQs on Element Material Balances and Answers

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Basic Chemical Engineering Multiple Choice Questions & Answers (MCQs) on “Element Material Balances”.

1. A reactor supplied with CO2 and H2O, the product contains 10% H2CO3 and 90% H2O, what is the feed rate of CO2, if the rate of product is 50 mole?
a) 5 mole
b) 10 mole
c) 25 mole
d) 50 mole
Answer: a
Clarification: The reaction is CO2 + H2O -> H2CO3. Let the rates of CO2 and H2O be F and W respectively. Element balances, C: F(1) = 0.1P(1), => 10F = P, => F = 5 mole.

2. A reactor supplied with CO2 and H2O, the product contains H2O and H2CO3, if the rate of CO2 is 10 mole, what is the percentage of H2CO3 in products?
a) 10
b) 50
c) 75
d) 100
Answer: d
Clarification: The reaction is CO2 + H2O -> H2CO3. Let the rate of CO2 and H2O and product be F and W respectively, and the fraction of H2CO3 in products be x. Element balances, C: F(1) = x*10(1), => F = 10x, => x = 1, => percentage of H2CO3 = 100%.

3. A reactor supplied with CO2 and H2O, the product contains H2O and H2CO3, if the rate of H2O is 10 mole, what is the rate of products?
a) 5 mole
b) 10 mole
c) 15 mole
d) 20 mole
Answer: b
Clarification: The reaction is CO2 + H2O -> H2CO3. Let the rate of products be P, and the fraction of H2CO3 in products be x, material balances, H: 10(2) = x*P(2) + (1 – x)*P(2), => P = 10 mole.

4. A reactor supplied with CO2 and H2O, the product contains CO2, H2O and H2CO3, if the rate of products is 10 mole and percentage of CO2 in products is 20%, what is the feed rate of H2O?
a) 2 mole
b) 4 mole
c) 8 mole
d) 10 mole

Answer: c
Clarification: The reaction is CO2 + H2O -> H2CO3. Let the rate of H2O be W, and percentage of H2O in products be x. Material balances, H: W(2) = x*10(2) + (1 – x – 0.2)*10(2), => W = 8 mole.

5. A reactor supplied with CO2 and H2O, the product contains 40% CO2, 30% H2O and 30% H2CO3, what is the ratio of feed rate of CO2 and rate of products?
a) 0.3
b) 0.4
c) 0.7
d) 0.9
Answer: c
Clarification: The reaction is CO2 + H2O -> H2CO3. Let the rate of CO2, H2O and H2CO3 be F, W and P respectively. Element balances, C: F(1) = 0.4P(1) + 0.3P(1), => F/P = 0.7.

6. A reactor is supplied with Hexane and hydrogen to produce 20% methane, 30% propane and 50% butane, what is the ratio of hydrogen consumed and the hexane reacted?
a) 0.21
b) 0.45
c) 0.59
d) 0.73
Answer: d
Clarification: The reaction is 2C6H14 + 2H2 -> CH4 + 2C3H8 + C5H12, Let the rate of Hexane, hydrogen and products be F, W and P respectively. Element balances, C: 2F(6) = 0.2P(1) + 2*0.3P(3) + 0.5P(5), => 8F = 3P, H: 2F(14) + 2W(2) = 0.2P(4) + 2*0.3P(8) + 0.5P(12), => 7F + W = 2.9P, => W/F= 2.2/3 = 0.73.

7. A reactor is supplied with 10 mole of Hexane and some hydrogen to produce 20% methane, 30% propane and 50% butane, what is the rate of products?
a) 10 mole
b) 16.6 mole
c) 26.6 mole
d) 40 mole
Answer: c
Clarification: The reaction is 2C6H14 + 2H2 -> CH4 + 2C3H8 + C5H12, Let the rate of products be P. Element balance, C: 2*10(6) = 0.2P(1) + 2*0.3P(3) + 0.5P(5), => 120 = 4.5P, => P = 26.6 mole.

8. A reactor is supplied with glucose and oxygen to produce 40% CO2, 60% H2O, what is the ratio of glucose reacted and products?
a) 0.2
b) 0.4
c) 0.6
d) 0.8

Answer: b
Clarification: The reaction is C6H12O6 + 6O2 -> 6CO2 + 6H2O, Let the rate of glucose and products be F and P, Element balances, C: F(6) = 6*0.4P(1), => F/P = 0.4.

9. A reactor is supplied with 10 moles glucose and excess oxygen to produce 40% CO2, 60% H2O, what is the rate of products?
a) 25 mole
b) 50 mole
c) 75 mole
d) 100 mole
Answer: a
Clarification: The reaction is C6H12O6 + 6O2 -> 6CO2 + 6H2O, Let the rate of products be P. Element balances, C: 10(6) = 6*0.4P(1), => P = 25 mole.

10. A reactor is supplied with 10 moles glucose and excess oxygen to produce CO2 and H2O, if the rate of products is 25 moles, what is the percentage of CO2 in the products?
a) 0.1
b) 0.2
c) 0.3
d) 0.4
Answer: d
Clarification: The reaction is C6H12O6 + 6O2 -> 6CO2 + 6H2O. Let the fraction of CO2 be x, Element balances, C: 10(6) = 6*x*5(1), => x = 0.4.

11. Ethane is supplied at the rate F with oxygen at the rate W and leaves the following products CO2, H2O, O2, C2H6 and H2 at the rate P. How many independent element balances are possible?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: The reaction is C2H6 + 3.5O2 -> 2CO2 + 3H2O, since there are three elements C, H and O, we can write 3 independent material balances.

12. Ethane is supplied with chlorine in a reactor, how many independent element balances can you write?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: The reaction is C2H6 + Cl2 -> C2H5Cl + HCl, Element balance C: 2F = 2xP, H: 6F = 5xP + (1 – x)P, Cl: 2W = xP + (1 – x)P, => The equations are F = xP, 6F = 4xP + P, and 2W = P, which are independent.

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13. A reactor is supplied with 2 streams both at same rates, one has pure Cl2, and another has pure C2H6, if the product has 20% C2H5Cl and 80% HCl, what is the ratio of rate of feed and rate of product?
a) 0.2
b) 0.5
c) 2
d) 5
Answer: a
Clarification: The reaction is C2H6 + Cl2 -> C2H5Cl + HCl, Let the rate of feed and products be F and P respectively. Element balances, C: F(2) = 0.2P(2), => F/P = 0.2.

14. A reactor is supplied with 2 streams both at same rates, one has pure Cl2, and another has pure C2H6, if the product has 20% C2H5Cl and 50% HCl and 30% C2H6, what is the ratio of rate of feed and rate of product?
a) 0.2
b) 0.5
c) 2
d) 5
Answer: b
Clarification: The reaction is C2H6 + Cl2 -> C2H5Cl + HCl, Let the rate of feed and products be F and P respectively. Element balances, C: F(2) – 0.3P(2) = 0.2P(2), => F/P = 0.5.

15. A reactor is supplied with 2 streams both at the rate 5 mole, one has 40% Cl2 and 60% N2, and another has pure C2H6, if the product has C2H5Cl, HCl and N2 at the rate 20 mole, what is the percentage of HCl in products?
a) 10%
b) 15%
c) 25%
d) 30%
Answer: b
Clarification: The reaction is C2H6 + Cl2 -> C2H5Cl + HCl, Let the fraction of C2H5Cl, HCl and N2 be x, y and (1 – x – y) respectively. Element balances, Cl: 0.4*5(2) = x*10(1) + y*10(1), => x + y = 0.4, C: 5(2) = x*20(2), => x = 0.25, => y = 0.4 – 0.25 = 0.15, => percentage of HCl = 15%.

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