# Bioprocess Engineering Questions and Answers – Law of Conservation of Mass

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

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

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

Explanation: Mg (Magnesium) + 2 HCl (Hydrochloric acid) = MgCl2 (Magnesium chloride) + H2 (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, 103 kg cellulose and 105 kg bacteria enter in the feed stream, while 102 kg cellulose and 1.5 x 102 kg bacteria leave in the effluent. The rate of cellulose digestion by the bacteria is 8 x 102 kg d-1. The rate of bacterial growth is 4x 102 kg d-l; the rate of cell death by lysis is 6 x 102 kg d-1. Write balances for cellulose and bacteria in the system.
a) 1× 103 kg, 9 × 103 kg
b) 1× 102 kg, 9.965 × 104 kg
c) 1× 103 kg, 9.964 × 104 kg
d) 1× 102 kg, 9 × 103 kg

Explanation: Cellulose is not generated by the process, only consumed. Using a basis of 1 day, the cellulose balance in kg is :
(103 – 102 + 0 – 8 x 102) = accumulation
Therefore, 1× 102 kg cellulose accumulates in the system each day.
Performing the same balance for bacteria:
(105 – 1.5 x 102 + 4 x 102 – 6 x 102) = accumulation
Therefore, 9.965 × 104 kg bacterial cells accumulate in the system each day.
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5. Which of the following system follows the differential equation?
a) Semi- Batch process
b) Batch process
c) Fed- Batch process
d) Continuous process

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. 2SO2+O2 → 2SO3. What is the stoichiometric ratio of SO2 to SO3?
a) 3
b) 1
c) 2
d) 0

Explanation: stoichiometric ratio of SO2 to SO3 = (2 mole of SO2 reacted)/(2 mole of SO3 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

Explanation: Material Balances (Steady-State, Non-Reactive Process) :
Total Balance: m1 + m2 = m3
Methanol-Balance: m1xM1 + m2xM2 = m3xM3
Water-Balance: m1xw1 + m2xw2 = m3xw3

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

Physical Constraint (applied to mixture 3):
xM3 + xW3 = 1.00

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

Total Balance (m3)
↓
Methanol Balance (xM3)
↓
Physical Constraint (xW3)
Total balance:
m3 = (200 kg) + (150 kg)
m3 = 350 kg

CH3OH balance:
(200 kg)(0.40) + (150 kg)(0.70) = (350 g)xM3
xM3 = 0.529

Physical constraint:
xW3 = 1.00 – xM3 = 1 – 0.529
xW3 = 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/m3. 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 m3
b) 14.5%, 22.4%, 0.059, 2.88 moles in m3
c) 16.7%, 22.1%, 0.058, 3.77 moles in m3
d) 14.7%, 23.1%, 0.059, 2.77 moles in m3

Explanation: (i) Weight fraction:
20/(100+20) = 0.167; %weight/weight = 16.7%

(ii) Weight/volume:
A density of 1323kg/m3 means that lm3 of solution weighs 1323kg, but 1323kg of salt solution contains
(20×1323kg of salt) / (100+20) = 220.5 kg salt/m3
1 m3 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 m3.

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

Explanation: Ethanol (C2H5OH) = 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 (C3H8) burns in this reaction:
C3H8 + 5O2 = 4H2O + 3CO2
If 200 g of propane is burned, how many g of H2O is produced?
a) 327.27 g
b) 345.5 g
c) 323.2 g
d) 232.3 g

Explanation: Steps to getting this answer:

Since you cannot calculate from grams of reactant to grams of products you must convert from grams of C3H8 to moles of C3H8 then from moles of C3H8 to moles of H2O. Then convert from moles of H2O to grams of H2O.

Step 1: 200 g C3H8 is equal to 4.54 mol C3H8.
Step 2: Since there is a ratio of 4:1 H2O to C3H8, for every 4.54 mol C3H8 there are 18.18 mol H2O.
Step 3: Convert 18.18 mol H2O to g H2O. 18.18 mol H2O is equal to 327.27 g H2O.

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