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# Multiple choice question for engineering

## Set 1

1. The given numbers are shown with their significant figures respectively. Find out the wrong one option?
a) 26.24 and 4
b) 24000. and 5
c) 0.000460 and 3
d) 28000.0 and 5

Answer: d [Reason:] Zero after a decimal point counts in significant figure.

2. The theoretical production rate of a chemical is 1µg/(mL)(min.). What is the actual production rate of the chemical in lb mol/ (day)(ft3) with an efficiency of 98%?
a) 0.0880
b) 0.0800
c) 0.0449
d) 0.529

Answer: a [Reason:] 1 lb mole = 454 g mole and 1 L = 3.531×10-2 ft3.

3. Standard form of 2145.67 is
a) 2.14*103
b) 2.14567*103
c) 0.214*104
d) None of the mentioned

Answer: b [Reason:] 21456.7 = 2.14567*103.

4. Correctly rounded, the quotient 5.000 g / 50.0 mL is
a) 0.10 g/mL
b) 0.1 g/mL
c) 0.100 g/mL
d) 0.1000 g/mL

Answer: c [Reason:] The measurement with the least number of significant digits determines the number of significant digits in the quotient.

5. Correctly rounded, the product 3.000 cm × 30.0 cm is
a) 9.0 x 101 cm2
b) 9.000 x 101 cm2
c) 9 x 101 cm2
d) 9.00 x 101 cm2

Answer: [Reason:] The measurement with the least number of significant digits determines the number of significant digits in the quotient.

6. Number of digits that are reasonably sure, called significant figures.
Above given definition is
a) Correct
b) Incorrect
c) Probably correct
d) None of the mentioned

Answer: a [Reason:] Number of digits that are reasonably sure, called significant figures.

7. 0.009046879855, round off the number to three significant figures
a) 0.0905
b) 0.0904
c) 0.09046
d) 0.09035

Answer: a [Reason:] In such situation we round up the last digit from 4 up to 5.

8. How many significant numbers are there in 420?
a) 1
b) 2
c) 3
d) 0

Answer: b [Reason:] Zero does not count in significant number here.

9. Using the proper number of significant figures 6 X 6 equals to
a) 3
b) 4
c) 36
d) 40

Answer: d [Reason:] Significant figure in 40 is only one that is same as 6.

10. How many significant figures are there in 0.002000?
a) 1
b) 2
c) 3
d) 4

Answer: d [Reason:] Zero after the point counts.

## Set 2

1-5. For a given reaction
C2H6 + Cl2 → C2H5Cl + HCl
Assume that the percentage conversion of the limiting reactant is 60% and the feed composition in mole percent is 50% C2H6, 40% Cl2 and 10% N2.
1. What is the mole percent of C2H6 in the product?
a) 24
b) 26
c) 40
d) 60

Answer: b [Reason:] Take basis = 100 moles, Reacting moles in the reaction = 0.60*40 = 24 moles, moles left = 50 – 24 = 26.

2. What is the mole percent of Cl2 in the product?
a) 16
b) 24
c) 26
d) 40

Answer: a [Reason:] Moles left = 40-24 = 16.

3. What is the mole percent of C2H5Cl in the product?
a) 16
b) 24
c) 26
d) 40

Answer: b [Reason:] Moles of C2H5Cl produced in the reaction = 24 moles.

4. What is the mole percent of HCl in the product?
a) 16
b) 24
c) 26
d) 40

Answer: b [Reason:] Moles of HCl produced in the reaction = 24 moles.

5. What is the mole percent of N2 in the product?
a) 10
b) 16
c) 24
d) 40

Answer: a [Reason:] N2 is not involved in the reaction, Moles of N2 = 10 moles.

6-10. In a bioreactor, these following reaction can occur to perform the fermentation process-
C6H12O6 → 2C2H5OH + 2CO2
C6H12O6 → 2C2H3CO2H + 2H2O
The tank is initially charged with 6000 kg of 15% solution of glucose in water. After fermentation, 160 kg of CO2 produced and 100 kg of unreacted glucose left in the broth.

6. How much water is there in the tank after fermentation?
a) 94.45 kg
b) 5100 kg
c) 5194.45 kg
d) None of the mentioned

Answer: c [Reason:] Species mole balance for each.

7. How many moles of C6H12O6 reacted in the first reaction?
a) 1.414
b) 1.818
c) 2.424
d) 4.848

Answer: b [Reason:] Species mole balance for each.

8. How much Ethanol is there in the product in kgs?
a) 67.43
b) 167.43
c) 267.43
d) 367.43

Answer: b [Reason:] Species mole balance for each.

9. How much propenoic acid is there in the product in kgs?
a) 178
b) 278
c) 378
d) 478

Answer: c [Reason:] Species mole balance.

10. How many kgs of C6H12O6 is used in second reaction?
a) 127.54
b) 227.54
c) 327.54
d) 427.54

Answer: d [Reason:] Species mole balance.

## Set 3

1-5. For a given reaction
C2H6 + Cl2 → C2H5Cl + HCl
Assume that the percentage conversion of the limiting reactant is 60% and the feed composition in mole percent is 50% C2H6, 40% Cl2 and 10% N2.
1. What is the mole percent of C2H6 in the product?
a) 24
b) 26
c) 40
d) 60

Answer: b [Reason:] Take basis = 100 moles, Reacting moles in the reaction = 0.60*40 = 24 moles, moles left = 50 – 24 = 26.

2. What is the mole percent of Cl2 in the product?
a) 16
b) 24
c) 26
d) 40

Answer: a [Reason:] Moles left = 40-24 = 16.

3. What is the mole percent of C2H5Cl in the product?
a) 16
b) 24
c) 26
d) 40

Answer: b [Reason:] Moles of C2H5Cl produced in the reaction = 24 moles.

4. What is the mole percent of HCl in the product?
a) 16
b) 24
c) 26
d) 40

Answer: b [Reason:] Moles of HCl produced in the reaction = 24 moles.

5. What is the mole percent of N2 in the product?
a) 10
b) 16
c) 24
d) 40

Answer: a [Reason:] N2 is not involved in the reaction, Moles of N2 = 10 moles.

6-10. In a bioreactor, these following reaction can occur to perform the fermentation process-
C6H12O6 → 2C2H5OH + 2CO2
C6H12O6 → 2C2H3CO2H + 2H2O
The tank is initially charged with 6000 kg of 15% solution of glucose in water. After fermentation, 160 kg of CO2 produced and 100 kg of unreacted glucose left in the broth.

6. How much water is there in the tank after fermentation?
a) 94.45 kg
b) 5100 kg
c) 5194.45 kg
d) None of the mentioned

Answer: c [Reason:] Species mole balance for each.

7. How many moles of C6H12O6 reacted in the first reaction?
a) 1.414
b) 1.818
c) 2.424
d) 4.848

Answer: b [Reason:] Species mole balance for each.

8. How much Ethanol is there in the product in kgs?
a) 67.43
b) 167.43
c) 267.43
d) 367.43

Answer: b [Reason:] Species mole balance for each.

9. How much propenoic acid is there in the product in kgs?
a) 178
b) 278
c) 378
d) 478

Answer: c [Reason:] Species mole balance.

10. How many kgs of C6H12O6 is used in second reaction?
a) 127.54
b) 227.54
c) 327.54
d) 427.54

Answer: d [Reason:] Species mole balance.

## Set 4

1. The given statement for stoichiometry is
“Stoichiometry provides a quantitative means of relating the amount of products produced by a chemical reaction to amount of reaction and vice versa.”
a) True
b) False
c) Insufficient information to predict
d) None of the mentioned

Answer: a [Reason:] The above given statement is a correct definition of stoichiometry.

2. The law of _____________ states that chemical reactions proceed with fix ratios of the number of reactant and products and products involved in the reaction.
a) Mass Conservation
b) Mole Conservation
c) Constant Proportionality
d) None of the mentioned

Answer: c [Reason:] The law of constant proportionality states that chemical reactions proceed with fix ratios of the number of reactant and products and products involved in the reaction.

3-5. C6H12O6 + a O2 → b CO2 + c H2O
For the above given reaction
3. What is the value of a?
a) 2
b) 4
c) 6
d) 8

Answer: c [Reason:] Oxygen, Carbon & Hydrogen Balance.

4. What is the value of b?
a) 2
b) 4
c) 6
d) 8

Answer: c [Reason:] Oxygen, Carbon & Hydrogen Balance.

5. What is the value of c?
a) 2
b) 4
c) 6
d) 8

Answer: c [Reason:] Oxygen, Carbon & Hydrogen Balance.

For the given balanced reaction
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O (Mol. Wt. of Glucose = 180)
In the below given questions if there is not specified the amount of a component then consider that in excess.
6. For 360 gms of Glucose how many gms of O2 is required?
a) 124
b) 284
c) 308
d) 384

Answer: d [Reason:] 1 mole of Glucose = 6 moles of O2.

7. For producing 6 mole of H2O, how many Kgs of Glucose is required?
a) 180
b) 360
c) 540
d) 720

Answer: a [Reason:] 1 mole of Glucose = 6 moles of H2O.

8. If the amount of CO2 produced is 44 gms, then How many Kgs of Glucose was in the feed?
a) 30
b) 60
c) 90
d) 120

Answer: a [Reason:] 1 mole of Glucose = 6 moles of CO2.

9. By taking 360 gms of Glucose how many gms of CO2 can be produced?
a) 228
b) 328
c) 428
d) 528

Answer: d [Reason:] 1 mole of Glucose = 6 moles of CO2.

10. By taking 64 gms of O2, How many gms of H2O can be produced?
a) 18
b) 36
c) 54
d) 72

Answer: b [Reason:] 1 mole of H2O = 61moles of O2.

## Set 5

1. The given statement for stoichiometry is
“Stoichiometry provides a quantitative means of relating the amount of products produced by a chemical reaction to amount of reaction and vice versa.”
a) True
b) False
c) Insufficient information to predict
d) None of the mentioned

Answer: a [Reason:] The above given statement is a correct definition of stoichiometry.

2. The law of _____________ states that chemical reactions proceed with fix ratios of the number of reactant and products and products involved in the reaction.
a) Mass Conservation
b) Mole Conservation
c) Constant Proportionality
d) None of the mentioned

Answer: c [Reason:] The law of constant proportionality states that chemical reactions proceed with fix ratios of the number of reactant and products and products involved in the reaction.

3-5. C6H12O6 + a O2 → b CO2 + c H2O
For the above given reaction
3. What is the value of a?
a) 2
b) 4
c) 6
d) 8

Answer: c [Reason:] Oxygen, Carbon & Hydrogen Balance.

4. What is the value of b?
a) 2
b) 4
c) 6
d) 8

Answer: c [Reason:] Oxygen, Carbon & Hydrogen Balance.

5. What is the value of c?
a) 2
b) 4
c) 6
d) 8

Answer: c [Reason:] Oxygen, Carbon & Hydrogen Balance.

For the given balanced reaction
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O (Mol. Wt. of Glucose = 180)
In the below given questions if there is not specified the amount of a component then consider that in excess.
6. For 360 gms of Glucose how many gms of O2 is required?
a) 124
b) 284
c) 308
d) 384

Answer: d [Reason:] 1 mole of Glucose = 6 moles of O2.

7. For producing 6 mole of H2O, how many Kgs of Glucose is required?
a) 180
b) 360
c) 540
d) 720

Answer: a [Reason:] 1 mole of Glucose = 6 moles of H2O.

8. If the amount of CO2 produced is 44 gms, then How many Kgs of Glucose was in the feed?
a) 30
b) 60
c) 90
d) 120

Answer: a [Reason:] 1 mole of Glucose = 6 moles of CO2.

9. By taking 360 gms of Glucose how many gms of CO2 can be produced?
a) 228
b) 328
c) 428
d) 528