This set of Bioprocess Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Law of Conservation of Mass”.

1. Which is the correct order for the equation of mass balance of the system?

{A} – {B} + {C} – {D} = {Mass accumulated within system}

a) A. Mass consumed within system, B. Mass in through system boundaries, C. Mass out through system boundaries, D. Mass generated within system

b) A. Mass generated within system, B. Mass in through system boundaries, C. Mass out through system boundaries, D. Mass consumed within system

c) A. Mass in through system boundaries, B. Mass out through system boundaries, C. Mass generated within system, D. Mass consumed within system

d) A. Mass consumed within system, B. Mass generated within system, C. Mass out through system boundaries, D. Mass in through system boundaries

View Answer

Explanation: A mass balance for the system can be written in a general way to account for these possibilities:

{Mass in through system boundaries} – {Mass out through system boundaries} + {Mass generated within system} – {Mass consumed within system} = {Mass accumulated within system}.

2. If the Reactant iron is combined with Reactant Sulfur to form the product Iron sulfide, then what will be the atomic mass of the product?

a) 88

b) 44

c) 22

d) 80

View Answer

Explanation: Fe (iron) + S (Sulfur) = FeS (Iron sulfide),

Atomic masses: Fe = 56, S = 32

One atom of each element on each side of the equations

Law of conservation of mass balance: 56 + 32 = 88.

3. If the reactant Magnesium reacts with Reactant Hydrochloric acid to form the product Magnesium chloride + Hydrogen, then what will be the atomic mass of both the product?

a) 90

b) 97

c) 80

d) 87

View Answer

Explanation: Mg (Magnesium) + 2 HCl (Hydrochloric acid) = MgCl

_{2}(Magnesium chloride) + H

_{2}(Hydrogen),

Atomic masses: Mg = 24, H = 1, Cl = 35.5

one atom of Mg, 2 atoms of H and 2 atoms of Cl on both sides of the equation,

Law of conservation of mass balance: 24 + 2 × (1+ 35.5) = 24+ (2 × 35.5) + (2×1) = 97 (Both equal 97)

Note the subscript 2 after the Cl in magnesium chloride or the 2 after the H in the hydrogen molecule, means two atoms of that element.

The 2 before the HCl doubles the number of hydrochloric acid molecules.

4. A continuous process is set up for treatment of wastewater. Each day, 10^{3} kg cellulose and 10^{5} kg bacteria enter in the feed stream, while 10^{2} kg cellulose and 1.5 x 10^{2} kg bacteria leave in the effluent. The rate of cellulose digestion by the bacteria is 8 x 10^{2} kg d^{-1}. The rate of bacterial growth is 4x 10^{2} kg d^{-l}; the rate of cell death by lysis is 6 x 10^{2} kg d^{-1}. Write balances for cellulose and bacteria in the system.

a) 1× 10^{3} kg, 9 × 10^{3} kg

b) 1× 10^{2} kg, 9.965 × 10^{4} kg

c) 1× 10^{3} kg, 9.964 × 10^{4} kg

d) 1× 10^{2} kg, 9 × 10^{3} kg

View Answer

Explanation: Cellulose is not generated by the process, only consumed. Using a basis of 1 day, the cellulose balance in kg is :

(10

^{3}– 10

^{2}+ 0 – 8 x 10

^{2}) = accumulation

Therefore, 1× 10

^{2}kg cellulose accumulates in the system each day.

Performing the same balance for bacteria:

(10

^{5}– 1.5 x 10

^{2}+ 4 x 10

^{2}– 6 x 10

^{2}) = accumulation

Therefore, 9.965 × 10

^{4}kg bacterial cells accumulate in the system each day.

5. Which of the following system follows the differential equation?

a) Semi- Batch process

b) Batch process

c) Fed- Batch process

d) Continuous process

View Answer

Explanation: Amounts of mass entering and leaving the system are specified using flow rates, a mass balance based on rates is called a differential balance. Whereas, each term of the mass-balance equation in this case is a quantity of mass, not a rate. This type of balance is called an integral balance. Differential balances for continuous systems operating at steady state and integral balances for batch and semi-batch systems between initial and final states are used.

6. 2SO_{2}+O_{2} → 2SO_{3}. What is the stoichiometric ratio of SO_{2} to SO_{3}?

a) 3

b) 1

c) 2

d) 0

View Answer

Explanation: stoichiometric ratio of SO

_{2}to SO

_{3}= (2 mole of SO

_{2}reacted)/(2 mole of SO

_{3}produced) = 1.

7. Two methanol-water mixtures are contained in separate tanks. The first mixture contains 40.0 wt% methanol and the second contains 70.0 wt% methanol. If 200 kg of the first mixture is combined with 150 kg of the second, what are the mass and composition of the product (Total balance, Methanol balance, Physical constraint)?

a) 350 kg, 0.529, 0.471

b) 320 kg, 0.529, 0.521

c) 360 kg, 0.620, 0.471

d) 350 kg, 0.690, 0.760

View Answer

Explanation: Material Balances (Steady-State, Non-Reactive Process) :

Total Balance: m

_{1}+ m

_{2}= m

_{3}

Methanol-Balance: m

_{1}x

_{M1}+ m

_{2}x

_{M2}= m

_{3}x

_{M3}

Water-Balance: m

_{1}x

_{w1}+ m

_{2}x

_{w2}= m

_{3}x

_{w3}

(choose only 2 equations since one of them is no longer independent)

Physical Constraint (applied to mixture 3):

x_{M3} + x_{W3} = 1.00

Always start with the equation with the least number of unknowns if possible and minimize solving equations simultaneously.

Total Balance (m_{3})

↓

Methanol Balance (x_{M3})

↓

Physical Constraint (x_{W3})

Total balance:

m_{3} = (200 kg) + (150 kg)

m_{3} = 350 kg

CH_{3}OH balance:

(200 kg)(0.40) + (150 kg)(0.70) = (350 g)x_{M3}

x_{M3} = 0.529

Physical constraint:

x_{W3} = 1.00 – x_{M3} = 1 – 0.529

x_{W3} = 0.471.

8. A solution of common salt in water is prepared by adding 20 kg of salt to 100 kg of water, to make a liquid of density 1323 kg/m^{3}. Calculate the concentration of salt in this solution as a (a) weight fraction, (b) weight/volume fraction, (c) mole fraction, (d) molal concentration.

a) 15.7%, 23.1%, 0.058, 3.88 moles in m^{3}

b) 14.5%, 22.4%, 0.059, 2.88 moles in m^{3}

c) 16.7%, 22.1%, 0.058, 3.77 moles in m^{3}

d) 14.7%, 23.1%, 0.059, 2.77 moles in m^{3}

View Answer

Explanation: (i) Weight fraction:

20/(100+20) = 0.167; %weight/weight = 16.7%

(ii) Weight/volume:

A density of 1323kg/m^{3} means that lm^{3} of solution weighs 1323kg, but 1323kg of salt solution contains

(20×1323kg of salt) / (100+20) = 220.5 kg salt/m^{3}

1 m^{3} solution contains 220.5 kg salt.

Weight/volume fraction = 220.5 / 1000 = 0.2205

And so weight / volume = 22.1%

(iii) Moles of water = 100 / 18 = 5.56

Moles of salt = 20 / 58.5 = 0.34

Mole fraction of salt = 0.34 / (5.56 + 0.34) = 0.058

(iv) The molar concentration (M) is 220.5/58.5 = 3.77 moles in m^{3}.

9. Calculate the degree of reduction for ethanol, where the degree of reduction of C=4, H=1, O=-2.

a) γ = 9

b) γ = 6

c) γ = 3

d) γ = 2

View Answer

Explanation: Ethanol (C

_{2}H

_{5}OH) = 2(4) + 6(1) + 1(-2) = 12,

γ = 12/2 = 6

As, the number of equivalents of available electrons per gram atom C is the measure of degree of reduction.

10. Propane (C_{3}H_{8}) burns in this reaction:

C_{3}H_{8} + 5O_{2} = 4H_{2}O + 3CO_{2}

If 200 g of propane is burned, how many g of H_{2}O is produced?

a) 327.27 g

b) 345.5 g

c) 323.2 g

d) 232.3 g

View Answer

Explanation: Steps to getting this answer:

Since you cannot calculate from grams of reactant to grams of products you must convert from grams of C_{3}H_{8} to moles of C_{3}H_{8} then from moles of C_{3}H_{8} to moles of H_{2}O. Then convert from moles of H_{2}O to grams of H_{2}O.

Step 1: 200 g C_{3}H_{8} is equal to 4.54 mol C_{3}H_{8}.

Step 2: Since there is a ratio of 4:1 H_{2}O to C_{3}H_{8}, for every 4.54 mol C_{3}H_{8} there are 18.18 mol H_{2}O.

Step 3: Convert 18.18 mol H_{2}O to g H_{2}O. 18.18 mol H_{2}O is equal to 327.27 g H_{2}O.

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