The general material balance equation for a process with reaction is

Input + Generation = Output + Consumption + Accumulation

Degree of freedom analysis in process calculations determines

how many variables must be specified to solve balances

Stoichiometric ratio in a reaction A → 2B means

1 mol A produces 2 mol B at complete conversion

Percent conversion is defined as

moles of reactant reacted divided by moles fed times 100

Chemical Process Calculations MCQs 2026

67 questions with detailed answers · 25 from past papers · 7 quiz batches available

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Page 1 of 1 Questions 1–10 of 67

Q1 medium

A stream of 425 kg/h contains 55 wt% key component. Component mass flow is
1. A 191.3 kg/h
2. B 233.8 kg/h
3. C 7.7 kg/h
4. D 0.1294 kg/h
💡 Explanation:
Mass fraction × total mass: 55% of 425.
Q2 Past Paper · PPSC/FPSC/CSS easy

The general material balance equation for a process with reaction is
1. A Input = Output only always
2. B Generation equals consumption always in batch
3. C Input + Generation = Output + Consumption + Accumulation
4. D Accumulation is always zero in unsteady state
💡 Explanation:
Component balance includes generation and consumption terms.
Q3 Past Paper · PPSC/FPSC/CSS medium

Degree of freedom analysis in process calculations determines
1. A only the colour of piping
2. B turbine blade angle only
3. C market share of product
4. D how many variables must be specified to solve balances
💡 Explanation:
DOF = unknowns − independent equations.
Q4 Past Paper · PPSC/FPSC/CSS easy

Stoichiometric ratio in a reaction A → 2B means
1. A 2 mol A produce 1 mol B
2. B mass of A equals mass of B
3. C 1 mol A produces 2 mol B at complete conversion
4. D volume always doubles regardless of gas law
💡 Explanation:
Mole ratio follows balanced equation coefficients.
Q5 Past Paper · PPSC/FPSC/CSS easy

Percent conversion is defined as
1. A moles product divided by moles feed
2. B mass catalyst divided by mass feed
3. C heat released divided by feed enthalpy
4. D moles of reactant reacted divided by moles fed times 100
💡 Explanation:
Conversion measures extent of reactant consumption.
Q6 medium

Selectivity in parallel reactions producing desired product D is
1. A total moles all products divided by feed
2. B profit divided by cost
3. C reactor volume divided by feed rate
4. D moles D formed divided by moles of key reactant consumed
💡 Explanation:
Selectivity quantifies preference for desired product.
Q7 Past Paper · PPSC/FPSC/CSS medium

Yield based on reactant is
1. A moles of desired product divided by moles of reactant fed
2. B moles product divided by moles reacted only without feed basis
3. C mass catalyst used divided by product mass
4. D energy out divided by energy in
💡 Explanation:
Yield ties product to original feed amount.
Q8 Past Paper · PPSC/FPSC/CSS easy

Recycle stream in a process is used to
1. A increase purge losses always
2. B bypass all separation units
3. C return unreacted material or solvent back to the feed
4. D eliminate need for stoichiometry
💡 Explanation:
Recycle improves raw material utilization.
Q9 medium

Bypass stream skips
1. A one or more processing units while main stream passes through
2. B all feed to atmosphere
3. C only energy balance terms
4. D the recycle compressor always
💡 Explanation:
Bypass adjusts composition or protects sensitive equipment.
Q10 Past Paper · PPSC/FPSC/CSS medium

Purge stream is withdrawn to
1. A increase inert concentration always
2. B replace product withdrawal
3. C heat the reactor
4. D prevent buildup of inerts or impurities in recycle loop
💡 Explanation:
Purge controls impurity accumulation in closed loops.
Q11 easy

Basis of 100 mol feed in stoichiometry simplifies
1. A only energy units conversion
2. B pipe wall thickness design
3. C pump NPSH calculation
4. D ratio calculations for extent of reaction
💡 Explanation:
Percentage compositions map directly from 100 mol basis.
Q12 Past Paper · PPSC/FPSC/CSS easy

Limiting reactant in a reaction mixture is
1. A the reactant that would be exhausted first if reaction went to completion
2. B the reactant present in largest mass always
3. C the catalyst pellet
4. D the solvent with highest boiling point
💡 Explanation:
Extent is limited by the stoichiometric deficient reactant.
Q13 medium

Excess reactant is often fed to
1. A drive conversion of limiting reactant toward completion
2. B reduce total plant cost always without benefit
3. C increase inert purge only
4. D avoid energy balance
💡 Explanation:
Excess shifts equilibrium and improves conversion in reversible reactions.
Q14 Past Paper · PPSC/FPSC/CSS medium

Dalton law is useful in gas mixture material balances when
1. A liquids form non-ideal solutions only
2. B solids dissolve completely always
3. C reaction is highly exothermic
4. D total pressure equals sum of partial pressures at same temperature
💡 Explanation:
Gas mole fractions relate to partial pressures for ideal gases.
Q15 medium

Raoult law relates vapor phase mole fraction to
1. A liquid mole fraction times pure component vapor pressure divided by total pressure
2. B only solid solubility
3. C reaction rate constant
4. D heat transfer coefficient
💡 Explanation:
y_i = x_i P_i^sat / P for ideal liquid mixtures.
Q16 Past Paper · PPSC/FPSC/CSS easy

Energy balance on a steady flow process is based on
1. A first law of thermodynamics for open systems
2. B second law only without enthalpy
3. C Newton viscosity law
4. D Darcy friction equation
💡 Explanation:
ΔH + ΔKE + ΔPE = Q − W_s (sign convention dependent).
Q17 easy

Sensible heat term in energy balance accounts for
1. A latent heat of vaporization only
2. B temperature change without phase change
3. C chemical reaction enthalpy only
4. D pump hydraulic power only
💡 Explanation:
Q_sensible = m Cp ΔT for single phase.
Q18 Past Paper · PPSC/FPSC/CSS medium

Latent heat at constant pressure appears when
1. A only solids are crushed
2. B only gases are compressed isothermally without phase change
3. C phase change occurs at nearly constant temperature
4. D only mixing of inerts
💡 Explanation:
Vaporization or condensation adds m λ term.
Q19 medium

Reference state for enthalpy calculations is
1. A arbitrary but must be used consistently
2. B always 25°C only by law
3. C always absolute zero for all tables
4. D always reactor inlet temperature
💡 Explanation:
Enthalpy is relative; consistency matters in balances.
Q20 Past Paper · PPSC/FPSC/CSS medium

Heat of reaction at constant pressure is represented by
1. A change in internal energy only always
2. B change in enthalpy of products minus reactants
3. C kinetic energy of fluids
4. D pump head
💡 Explanation:
ΔH_rxn from formation enthalpies or Hess law.
Q21 Past Paper · PPSC/FPSC/CSS easy

Hess law states that total enthalpy change for a reaction is
1. A always zero for exothermic steps
2. B equal to entropy change
3. C proportional to catalyst mass
4. D independent of path between same initial and final states
💡 Explanation:
State function property enables stepwise addition.
Q22 Past Paper · PPSC/FPSC/CSS hard

Orsat analysis in combustion calculations measures
1. A dry volumetric composition of flue gas
2. B liquid viscosity
3. C catalyst surface area
4. D pipe roughness
💡 Explanation:
CO2, O2, CO on dry basis for excess air calculation.
Q23 medium

Theoretical air for complete combustion of hydrocarbon is calculated from
1. A oxygen required by carbon and hydrogen in fuel
2. B only nitrogen in stack
3. C only sulfur in feedwater
4. D compressor polytropic exponent
💡 Explanation:
Stoichiometric O2 from C → CO2 and H → H2O.
Q24 Past Paper · PPSC/FPSC/CSS medium

A stream of 467 kg/h contains 47 wt% key component. Component mass flow is
1. A 247.5 kg/h
2. B 9.9 kg/h
3. C 219.5 kg/h
4. D 0.1006 kg/h
💡 Explanation:
Mass fraction × total mass: 47% of 467.
Q25 easy

A stream of 51 kg/h contains 61 wt% key component. Component mass flow is
1. A 19.9 kg/h
2. B 0.8 kg/h
3. C 1.1961 kg/h
4. D 31.1 kg/h
💡 Explanation:
Mass fraction × total mass: 61% of 51.
Q26 easy

A stream of 85 kg/h contains 75 wt% key component. Component mass flow is
1. A 21.3 kg/h
2. B 63.8 kg/h
3. C 1.1 kg/h
4. D 0.8824 kg/h
💡 Explanation:
Mass fraction × total mass: 75% of 85.
Q27 hard

Adiabatic reactor energy balance with no shaft work simplifies to
1. A Q is large always
2. B ΔPE dominates always
3. C kinetic terms always zero without check
4. D sum of enthalpy flows in equals sum out including reaction heat
💡 Explanation:
Adiabatic: Q=0.
Q28 Past Paper · PPSC/FPSC/CSS medium

A stream of 119 kg/h contains 89 wt% key component. Component mass flow is
1. A 13.1 kg/h
2. B 105.9 kg/h
3. C 1.3 kg/h
4. D 0.7479 kg/h
💡 Explanation:
Mass fraction × total mass: 89% of 119.
Q29 easy

Extent of reaction ξ relates moles reacted of limiting species by
1. A moles reacted = mass/ξ always
2. B ξ equals conversion percent directly without definition
3. C moles reacted = ν_i ξ
4. D ξ is always 1
💡 Explanation:
Stoichiometric extent links all species changes.
Q30 easy

A stream of 153 kg/h contains 23 wt% key component. Component mass flow is
1. A 117.8 kg/h
2. B 6.7 kg/h
3. C 0.1503 kg/h
4. D 35.2 kg/h
💡 Explanation:
Mass fraction × total mass: 23% of 153.
Q31 easy

A stream of 187 kg/h contains 37 wt% key component. Component mass flow is
1. A 117.8 kg/h
2. B 5.1 kg/h
3. C 69.2 kg/h
4. D 0.1979 kg/h
💡 Explanation:
Mass fraction × total mass: 37% of 187.
Q32 Past Paper · PPSC/FPSC/CSS hard

A stream of 221 kg/h contains 51 wt% key component. Component mass flow is
1. A 112.7 kg/h
2. B 108.3 kg/h
3. C 4.3 kg/h
4. D 0.2308 kg/h
💡 Explanation:
Mass fraction × total mass: 51% of 221.
Q33 easy

A stream of 255 kg/h contains 65 wt% key component. Component mass flow is
1. A 165.8 kg/h
2. B 89.3 kg/h
3. C 3.9 kg/h
4. D 0.2549 kg/h
💡 Explanation:
Mass fraction × total mass: 65% of 255.
Q34 Past Paper · PPSC/FPSC/CSS medium

If recycle ratio R is defined as recycle flow divided by fresh feed, increasing R with fixed purge usually
1. A eliminates all impurities instantly
2. B reduces need for separation always
3. C violates mass conservation
4. D increases impurity buildup unless purge increases
💡 Explanation:
Recycle concentrates species not removed.
Q35 easy

A stream of 289 kg/h contains 79 wt% key component. Component mass flow is
1. A 60.7 kg/h
2. B 3.7 kg/h
3. C 0.2734 kg/h
4. D 228.3 kg/h
💡 Explanation:
Mass fraction × total mass: 79% of 289.
Q36 easy

Combustion material balance on carbon gives moles CO2 produced equal to
1. A moles oxygen fed only
2. B moles nitrogen in air
3. C moles water formed only
4. D moles carbon in fuel reacted
💡 Explanation:
C + O2 → CO2: 1:1 mole basis.
Q37 Past Paper · PPSC/FPSC/CSS medium

A stream of 323 kg/h contains 13 wt% key component. Component mass flow is
1. A 281.0 kg/h
2. B 24.8 kg/h
3. C 42.0 kg/h
4. D 0.0402 kg/h
💡 Explanation:
Mass fraction × total mass: 13% of 323.
Q38 hard

Wet basis to dry basis conversion for moisture M% removes
1. A only ash content
2. B only heating value
3. C water mass fraction from total
4. D only nitrogen
💡 Explanation:
Dry basis excludes water.
Q39 easy

A stream of 357 kg/h contains 27 wt% key component. Component mass flow is
1. A 260.6 kg/h
2. B 96.4 kg/h
3. C 13.2 kg/h
4. D 0.0756 kg/h
💡 Explanation:
Mass fraction × total mass: 27% of 357.
Q40 Past Paper · PPSC/FPSC/CSS medium

Energy balance term for shaft work of compressor is
1. A always heat loss only
2. B work input to system (sign per convention)
3. C always potential energy only
4. D always reaction enthalpy
💡 Explanation:
Compressors add enthalpy via shaft work.
Q41 easy

A stream of 391 kg/h contains 41 wt% key component. Component mass flow is
1. A 230.7 kg/h
2. B 160.3 kg/h
3. C 9.5 kg/h
4. D 0.1049 kg/h
💡 Explanation:
Mass fraction × total mass: 41% of 391.
Q42 Past Paper · PPSC/FPSC/CSS medium

A stream of 263 kg/h contains 43 wt% key component. Component mass flow is
1. A 149.9 kg/h
2. B 6.1 kg/h
3. C 113.1 kg/h
4. D 0.1635 kg/h
💡 Explanation:
Mass fraction × total mass: 43% of 263.
Q43 easy

A stream of 213 kg/h contains 73 wt% key component. Component mass flow is
1. A 57.5 kg/h
2. B 155.5 kg/h
3. C 2.9 kg/h
4. D 0.3427 kg/h
💡 Explanation:
Mass fraction × total mass: 73% of 213.
Q44 medium

A stream of 179 kg/h contains 59 wt% key component. Component mass flow is
1. A 73.4 kg/h
2. B 3.0 kg/h
3. C 105.6 kg/h
4. D 0.3296 kg/h
💡 Explanation:
Mass fraction × total mass: 59% of 179.
Q45 easy

A stream of 145 kg/h contains 45 wt% key component. Component mass flow is
1. A 79.8 kg/h
2. B 3.2 kg/h
3. C 65.3 kg/h
4. D 0.3103 kg/h
💡 Explanation:
Mass fraction × total mass: 45% of 145.
Q46 easy

A stream of 111 kg/h contains 31 wt% key component. Component mass flow is
1. A 76.6 kg/h
2. B 34.4 kg/h
3. C 3.6 kg/h
4. D 0.2793 kg/h
💡 Explanation:
Mass fraction × total mass: 31% of 111.
Q47 easy

Mole fraction of component i in mixture is
1. A moles i divided by total moles
2. B mass i divided by total mass always called mole fraction
3. C partial pressure only
4. D volume percent without conversion
💡 Explanation:
y_i or x_i = n_i/Σn.
Q48 medium

A stream of 77 kg/h contains 17 wt% key component. Component mass flow is
1. A 63.9 kg/h
2. B 4.5 kg/h
3. C 13.1 kg/h
4. D 0.2208 kg/h
💡 Explanation:
Mass fraction × total mass: 17% of 77.
Q49 easy

A stream of 493 kg/h contains 83 wt% key component. Component mass flow is
1. A 83.8 kg/h
2. B 5.9 kg/h
3. C 409.2 kg/h
4. D 0.1684 kg/h
💡 Explanation:
Mass fraction × total mass: 83% of 493.
Q50 hard

A stream of 459 kg/h contains 69 wt% key component. Component mass flow is
1. A 142.3 kg/h
2. B 6.7 kg/h
3. C 316.7 kg/h
4. D 0.1503 kg/h
💡 Explanation:
Mass fraction × total mass: 69% of 459.
Q51 hard

A stream of 433 kg/h contains 33 wt% key component. Component mass flow is
1. A 290.1 kg/h
2. B 13.1 kg/h
3. C 0.0762 kg/h
4. D 142.9 kg/h
💡 Explanation:
Mass fraction × total mass: 33% of 433.
Q52 easy

A stream of 399 kg/h contains 19 wt% key component. Component mass flow is
1. A 323.2 kg/h
2. B 21.0 kg/h
3. C 0.0476 kg/h
4. D 75.8 kg/h
💡 Explanation:
Mass fraction × total mass: 19% of 399.
Q53 Past Paper · PPSC/FPSC/CSS medium

A stream of 365 kg/h contains 85 wt% key component. Component mass flow is
1. A 54.8 kg/h
2. B 310.3 kg/h
3. C 4.3 kg/h
4. D 0.2329 kg/h
💡 Explanation:
Mass fraction × total mass: 85% of 365.
Q54 easy

Material balance on a mixer with two inlet streams requires
1. A sum of component flows in equals outlet component flow
2. B only energy terms
3. C only momentum terms
4. D no balance equations
💡 Explanation:
Steady mixer: Σn_in = n_out per component.
Q55 easy

A stream of 331 kg/h contains 71 wt% key component. Component mass flow is
1. A 96.0 kg/h
2. B 235.0 kg/h
3. C 4.7 kg/h
4. D 0.2145 kg/h
💡 Explanation:
Mass fraction × total mass: 71% of 331.
Q56 easy

A stream of 297 kg/h contains 57 wt% key component. Component mass flow is
1. A 169.3 kg/h
2. B 127.7 kg/h
3. C 5.2 kg/h
4. D 0.1919 kg/h
💡 Explanation:
Mass fraction × total mass: 57% of 297.
Q57 easy

Overall mass balance in a process with vent and purge still requires
1. A ignoring vent streams
2. B only liquid products
3. C only catalyst mass
4. D accounting for all outlet streams including losses
💡 Explanation:
All exits must be summed.
Q58 easy

Excess air in combustion is expressed as
1. A ratio of CO2 to N2 only
2. B flue gas dew point only
3. C percent above stoichiometric air supplied
4. D boiler steam pressure
💡 Explanation:
Excess air lowers flame temperature but aids complete burn.
Q59 Past Paper · PPSC/FPSC/CSS easy

A steady-state material balance without chemical reaction states that
1. A energy in equals energy out only
2. B moles always decrease
3. C volume in equals mass out
4. D total mass in equals total mass out
💡 Explanation:
Mass is conserved in non-reactive steady processes.
Q60 easy

A stream of 229 kg/h contains 29 wt% key component. Component mass flow is
1. A 162.6 kg/h
2. B 7.9 kg/h
3. C 66.4 kg/h
4. D 0.1266 kg/h
💡 Explanation:
Mass fraction × total mass: 29% of 229.
Q61 hard

A stream of 195 kg/h contains 15 wt% key component. Component mass flow is
1. A 29.3 kg/h
2. B 165.8 kg/h
3. C 13.0 kg/h
4. D 0.0769 kg/h
💡 Explanation:
Mass fraction × total mass: 15% of 195.
Q62 easy

Heat capacity Cp used in sensible heat calculation is typically
1. A only latent heat
2. B mass or molar heat capacity at constant pressure
3. C only thermal conductivity
4. D only diffusivity
💡 Explanation:
Q = m Cp ΔT.
Q63 Past Paper · PPSC/FPSC/CSS medium

A stream of 161 kg/h contains 81 wt% key component. Component mass flow is
1. A 30.6 kg/h
2. B 130.4 kg/h
3. C 2.0 kg/h
4. D 0.5031 kg/h
💡 Explanation:
Mass fraction × total mass: 81% of 161.
Q64 easy

A stream of 127 kg/h contains 67 wt% key component. Component mass flow is
1. A 41.9 kg/h
2. B 1.9 kg/h
3. C 0.5276 kg/h
4. D 85.1 kg/h
💡 Explanation:
Mass fraction × total mass: 67% of 127.
Q65 easy

A stream of 93 kg/h contains 53 wt% key component. Component mass flow is
1. A 43.7 kg/h
2. B 49.3 kg/h
3. C 1.8 kg/h
4. D 0.5699 kg/h
💡 Explanation:
Mass fraction × total mass: 53% of 93.
Q66 Past Paper · PPSC/FPSC/CSS medium

A stream of 59 kg/h contains 39 wt% key component. Component mass flow is
1. A 36.0 kg/h
2. B 1.5 kg/h
3. C 23.0 kg/h
4. D 0.6610 kg/h
💡 Explanation:
Mass fraction × total mass: 39% of 59.
Q67 easy

A stream of 475 kg/h contains 25 wt% key component. Component mass flow is
1. A 118.8 kg/h
2. B 356.3 kg/h
3. C 19.0 kg/h
4. D 0.0526 kg/h
💡 Explanation:
Mass fraction × total mass: 25% of 475.

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