This area carries 10–15 questions, more than any other, and its range grew in the 2020 outline. Almost every problem is the same accounting done carefully: pick a boundary and a basis, then make what goes in, comes out, accumulates and reacts add up.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Chemical
questions on the exam10–15
problems on this page18
of 17 knowledge areas08
Checked
What the specification covers
The July 2020 outline lists six subtopics. In our words:
Steady-state mass balances: mixers, splitters, separators and multi-unit flowsheets with no accumulation.
Unsteady-state mass balances: filling, draining and well-mixed tanks whose contents change with time.
Steady-state energy balances: heaters, coolers, evaporators and adiabatic mixing, with enthalpies from tables.
Unsteady-state energy balances: a tank or batch vessel heating up or cooling down.
Recycle and bypass: loops that change what the reactor sees compared with what the plant takes in.
Reactive systems: extents of reaction, limiting and excess reactants, combustion.
18 balance problems, worked out
0 / 18 answered · 0 correct
Question 1 of 18 · Material/Energy Balances · 10–15 items
An entire multi-unit chemical facility operates at steady state. A macro-scale boundary energy closure check is performed to estimate the unmeasured environmental heat loss (Qloss) to the surroundings. The plant has two primary feed streams and two primary product streams. The table below provides the mass flow rates and specific enthalpies for all streams crossing the facility's control boundary. The plant consumes 500 kW of electrical power to run internal pumps and compressors, all of which is dissipated as heat within the process fluid or lost to the environment. Determine the total environmental heat loss rate (Qloss) from the entire plant.
1360.0 kW leaves the waste stream D (450 kW) out of the outlet enthalpy.
1410.0 kW counts the 500 kW of electrical power twice.
Question 2 of 18 · Material/Energy Balances · 10–15 items
A process has a bypass stream where 30% of the feed is diverted around a reactor. The feed contains 50 kg/h of component A, and the reactor converts 80% of A to product B. How much component A leaves in the final product stream after the bypass rejoins the reactor outlet?
7 kg/h counts only the unconverted A leaving the reactor and forgets the 15 kg/h that bypassed it.
10 kg/h ignores the bypass and sends all 50 kg/h through the reactor (0.2 × 50).
15 kg/h counts only the bypassed A; the reactor outlet still carries 7 kg/h of unconverted A.
Bypass carries 0.30 × 50 = 15 kg/h of A; the reactor gets 35 kg/h and leaves 20 % = 7 kg/h; total 22 kg/h.
Question 3 of 18 · Material/Energy Balances · 10–15 items
In a steady-state evaporation process, a brine solution containing 15% salt is fed at 1200 kg/h and is concentrated to 40% salt by evaporating water. What is the rate of water (in kg/h) that must be evaporated?
Given
Feed rate = 1200 kg/h
Feed salt fraction = 15%
Product salt fraction = 40%
Find
Water evaporated (kg/h)
Solution
salt: 0.15 × 1,200 kg/h = 0.40·P → P = 450 kg/h
water evaporated = F − P = 1,200 − 450 = 750 kg/h
Salt is conserved: 0.15 × 1,200 = 180 kg/h = 0.40P, so P = 450 kg/h and water evaporated = 1,200 − 450 = 750 kg/h.
450 kg/h is the concentrated product rate (180 kg/h salt ÷ 0.40), not the water removed.
300 kg/h multiplies the feed by the difference in salt fractions (0.40 − 0.15); the salt balance does not work that way.
180 kg/h is the salt flow (0.15 × 1,200), the tie component, not the water evaporated.
Question 4 of 18 · Material/Energy Balances · 10–15 items
A steady-state chemical process features a primary reactor and a downstream separator. Unreacted materials are recovered in the separator and recycled back to mix with the fresh feed before entering the reactor. An engineering intern reviews the process and calculates the overall system mass balance by equating the final product flow rate to the sum of the fresh feed flow rate and the internal recycle stream flow rate. What fundamental process balance error is present in this control volume analysis?
Solution
Overall boundary (steady state): fresh feed = products; internal streams such as the recycle cancel out. The recycle appears only in balances around the mixer, reactor or separator.
Around the whole plant the recycle never crosses the boundary: fresh feed = product (+ purge). Adding the recycle counts an internal stream as an input.
For an overall total-mass balance generation is zero anyway; the mistake is in which streams are counted.
Mass balances do not need constant density; density matters only for volumetric flows.
The plant is an open, continuous system, and treating it as one is correct.
No inert or purge issue is described; at steady state there is no accumulation term at all.
Question 5 of 18 · Material/Energy Balances · 10–15 items
An engineer designs a steady-state reactor system featuring a completely closed recycle loop to recover 100% of the unreacted feed. The fresh feed contains 99% reactant and 1% trace inert gas. Which operational outcome will eventually occur if this closed-loop system operates continuously without modification?
Solution
inert balance: d(nI)/dt = 0.01·F − 0 > 0
Without an outlet the inert inventory grows without bound; a purge stream sets the steady state (purge·xI,loop = 0.01·F).
In − out = accumulation: inerts enter with every mole of feed but have no outlet, so they build up until the loop is purged or the process stops.
A steady state needs inerts to leave as fast as they enter; with a fully closed loop they cannot leave, so no steady state exists.
Inerts dilute the reactant and lower its partial pressure; they slow the reaction rather than speed it up.
1 % is the feed concentration; with no outlet the loop concentration climbs far above it.
Nothing balances the inert input; it keeps entering at 1 % of the feed with no exit.
Question 6 of 18 · Material/Energy Balances · 10–15 items
A steady-state chemical plant processes reactant A through a continuous reactor and a subsequent separator that recycles unreacted A. Given the absolute molar flow rates of reactant A measured at various strategic points in the system (detailed in the table below), calculate the overall conversion and the single-pass conversion of the reactor, respectively.
40 % is the single-pass value put first, and 66.7 % divides the reacted A by the reactor outlet; the order asked is overall, then single-pass.
Overall: (200 − 30)/200 = 85 % of the fresh feed; single pass: (425 − 255)/425 = 40 % of the reactor feed.
52.9 % is the recycle's share of the reactor feed (225/425), not a conversion; the single-pass value is (425 − 255)/425.
66.7 % divides the reacted A (170) by the reactor outlet (255) instead of the reactor inlet (425).
92.9 % divides (reactor inlet − product)/reactor inlet = 395/425; overall conversion is based on the fresh feed, (200 − 30)/200 = 85 %.
Question 7 of 18 · Material/Energy Balances · 10–15 items
A steady-state catalytic reactor converts exactly 20% of the reactant fed into it per pass. To prevent byproduct buildup, a portion of the unreacted stream is continuously purged, while the remainder is recycled to mix with the pure fresh feed. The process must maintain a constant consumption of 100 kmol/h of reactant. Initially, the purge fraction is 10% of the unreacted stream. If the operational purge fraction is abruptly doubled to 20%, calculate the necessary increase in the fresh feed rate required to maintain the same production output.
Reactor feed = 100/0.20 = 500 kmol/h, so 400 kmol/h is unreacted; fresh feed = 100 + purge = 140 at 10 % and 180 at 20 %, an increase of 40.0 kmol/h.
50.0 kmol/h applies the purge change to the whole reactor feed (500 kmol/h) instead of the unreacted stream.
80.0 kmol/h is the new purge flow (0.20 × 400 kmol/h unreacted), not the increase in fresh feed.
140.0 kmol/h is the original fresh feed at a 10 % purge, not the increase.
180.0 kmol/h is the new fresh feed at a 20 % purge, not the increase.
Question 8 of 18 · Material/Energy Balances · 10–15 items
Parameter
Value
Total feed flow rate
1,000 kg/h
Target water evaporation rate
400 kg/h
Feed inlet temperature
25 °C
Evaporator boiling temperature
95 °C
Feed specific heat capacity
3.8 kJ/(kg·°C)
Latent heat of vaporization
2,270 kJ/kg
A continuously operating evaporator processes a subcooled liquid feed. Heating steam must provide sufficient thermal energy to bring the feed up to its boiling point and to vaporize the necessary fraction of water. Using the specific property data provided in the table, determine the total heat requirement (in kW) that the steam must supply to the evaporator.
704.4 kW vaporizes the entire feed; only 400 kg/h of water evaporates.
Question 9 of 18 · Material/Energy Balances · 10–15 items
A continuous stirred-tank reactor (CSTR) with a constant volume V (in liters) initially contains pure water. At t = 0, a reactant feed stream with concentration C0 (in mol L−1) is introduced at a volumetric flow rate Q (in L min−1). The reactant undergoes a first-order decay with a rate constant k (in s−1). The effluent exits at the same volumetric flow rate Q. Which of the following differential equations correctly models the accumulation of reactant concentration C (in mol L−1) with respect to time t (in minutes)?
This converts the flow term to per second while the left side is per minute.
Multiplying the whole right side by 60 also inflates the flow term, which is already per minute.
Dividing k by 60 converts in the wrong direction: a per-second constant is 60 times larger per minute.
The flow term is per minute but k is per second, so the two terms on the right are in different time units.
With t in minutes, Q/V is in min−1 and k in s−1 must become 60k in min−1; this is the only dimensionally consistent form.
Question 10 of 18 · Material/Energy Balances · 10–15 items
Property
Value
Units
Cp,solid
2.11
kJ/(kg·K)
Cp,liquid
4.18
kJ/(kg·K)
Cp,vapor
2.08
kJ/(kg·K)
ΔHfus
334
kJ/kg
ΔHvap
2257
kJ/kg
Tmelt
0
°C
Tboil
100
°C
An engineer must calculate the total specific heat requirement (Δh, in kJ/kg) to convert water from ice at −15 °C to superheated steam at 150 °C at 1 atm. Using the constants in the table, which expression follows the correct path for this energy balance?
This uses the liquid cp for ice and steam as well; each phase has its own heat capacity (2.11 and 2.08).
This leaves out the heat of fusion (334 kJ/kg) needed to melt the ice at 0 °C.
Five steps in order: heat the ice 15 K, melt it, heat the liquid 100 K, vaporize it, then superheat the vapor 50 K, each with its own cp.
This heats the liquid water with the vapor heat capacity (2.08) instead of 4.18.
Only sensible terms: both latent heats, melting and vaporization, are missing.
Question 11 of 18 · Material/Energy Balances · 10–15 items
A steady-state reactor processes a continuous feed containing 10000 mol/h of A and 15000 mol/h of B. The reversible chemical reaction is A + 2B ⇌ 3C. The separated product stream contains 6000 mol/h of C. No component C is present in the initial feed. Utilizing the algebraic extent of reaction (ξ), what is the exiting molar flow rate of unreacted B?
9000 mol/h assumes one mole of B is used per mole of C formed; the stoichiometry is 2 B per 3 C.
3000 mol/h uses the 6,000 mol/h of C as the extent of reaction; ξ = 6,000/3 = 2,000 mol/h.
−21000 mol/h multiplies 6,000 by both coefficients (2 × 3); a negative flow is a sign the extent was taken wrongly.
Question 12 of 18 · Material/Energy Balances · 10–15 items
In an exothermic reactor, 100 kJ of heat is released per mole of A reacted. If A is fed at 2 mol/s and 80% conversion is achieved, how much heat is generated?
Given
Heat of reaction = 100 kJ released per mol of A reacted
40 kW uses the unconverted 20 % of the feed (2 × 0.20 × 100) instead of the 80 % that reacts.
80 kW leaves out the 2 mol/s feed rate: it is 100 kJ/mol × 0.80 per mole of feed.
100 kW takes the heat per mole as the total rate, ignoring both the feed rate and the conversion.
Moles of A reacted = 2 mol/s × 0.80 = 1.6 mol/s; heat released = 1.6 × 100 kJ/mol = 160 kW.
Question 13 of 18 · Material/Energy Balances · 10–15 items
A well-mixed chemical reactor holding liquid mass M with specific heat cp is heated by a coil supplying Q̇in while losing heat to the surroundings (Tamb) through a conductance UA. Liquid flows through at rate ṁ with inlet temperature Tin. Which equation gives the instantaneous rate of temperature change dT/dt?
This is the steady-state balance; the question asks for the instantaneous rate of change.
This drops the enthalpy carried by the flowing stream, as if the tank were a batch.
The flow term has the wrong sign: warmer inlet fluid (Tin > T) heats the tank.
The flow term lacks cp, and nothing divides by the liquid mass M, so the units do not work.
Mcp·dT/dt = ṁcp(Tin − T) + Q̇in − UA(T − Tamb); dividing by Mcp gives this form.
Question 14 of 18 · Material/Energy Balances · 10–15 items
Stream
Content
Density (kg/L)
ΔP (kPa)
1
Pure A
1.20
16.0
2
Pure B
0.80
36.0
A process perfectly mixes Stream 1 and Stream 2 to form Stream 3. Differential pressure (DP) orifice meters measure the flow on the inlet streams. The volumetric flow rate is related to DP by the equation Q = 0.50√(ΔP), where Q is in L/s and ΔP is in kPa. Assuming steady-state operation, calculate the total mass flow rate of Stream 3 in kg s−1.
Given
Orifice relation Q = 0.50·√ΔP (Q in L/s, ΔP in kPa)
2.4 kg/s is one stream only; streams 1 and 2 each carry 2.4 kg/s, and stream 3 is their sum.
Question 15 of 18 · Material/Energy Balances · 10–15 items
Stream
Phase
Mass Flow (kg/s)
Specific Enthalpy (kJ/kg)
Stream 1
Subcooled Liquid
2.0
420
Stream 2
Saturated Liquid
3.0
850
Stream 3
Superheated Steam
5.0
2800
Three steady-state water streams of differing phases are mixed adiabatically in a continuous process. Using the operational parameters provided in the table, calculate the specific enthalpy of the resulting mixed exit stream.
2800 kJ/kg is the steam's own enthalpy; mixing with liquid lowers it.
17390 is the total enthalpy flow in kW; it still has to be divided by the 10 kg/s total flow.
Question 16 of 18 · Material/Energy Balances · 10–15 items
A batch process is described by an unsteady material balance where the mass accumulation rate is dm/dt = R(t). At time t = 0, the system is charged with an initial mass m0. Which definite integral setup correctly yields the true system mass m(t) at any arbitrary time t > 0?
The limits mix a mass with a time, and subtracting m0 inside the integral has no basis.
The integration variable is time, so the lower limit must be t = 0, not the mass m0.
R(t) is the accumulation rate, so it is added; subtracting it reverses the sign of the balance.
Integrating dm/dt = R(t) from 0 to t gives m(t) − m0 = ∫R dτ, so m(t) = m0 + ∫0tR(τ)dτ.
Putting m0 inside the integral multiplies it by t; the initial mass is added once.
Question 17 of 18 · Material/Energy Balances · 10–15 items
A continuous mixer combines two streams: Stream A (100 kg/h, 20% solute by mass) and Stream B (150 kg/h, 40% solute by mass). What is the mass percentage of solute in the outlet stream?
Solute = 0.20 × 100 + 0.40 × 150 = 80 kg/h in 250 kg/h, so the outlet is 32 % solute.
30 % is the plain average of 20 % and 40 %; the streams are not equal in size.
28 % swaps the compositions between the streams; the larger stream B carries the 40 % solute.
24 % counts only stream B's solute (60 kg/h) over the 250 kg/h total, leaving out stream A's 20 kg/h.
Question 18 of 18 · Material/Energy Balances · 10–15 items
A process consists of two sequential treatment tanks operating at steady state. The first tank receives a 1200 kg/h feed containing 5% impurity. In the first tank, 40% of the impurity is removed. The effluent from the first tank enters a second tank where an additional 30% of the remaining impurity is removed. What is the overall impurity removal rate in kg/h from the initial feed?
60.0 kg/h is the impurity fed (0.05 × 1,200), as if the two tanks removed all of it.
42.0 kg/h adds the percentages (40 % + 30 % = 70 % of 60); the second tank only acts on what the first one left.
36.0 kg/h is the impurity left after the first tank, the feed to the second one.
Feed impurity = 0.05 × 1,200 = 60 kg/h; tank 1 removes 24 kg/h, tank 2 removes 0.30 × 36 = 10.8 kg/h, so 34.8 kg/h in total.
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Common traps
Mass basis against mole basis. Component balances in reacting systems close in moles of atoms or by extent, not in mass of species; a mass balance on species changes with reaction. Convert once, at the start.
Excess reactant from the wrong amount. Percent excess compares what is fed with what the limiting reactant needs, not with what reacted:
fractional excess = (nfed − nstoich)/nstoich
Too few equations. Count unknowns against independent balances before calculating; a flowsheet with one degree of freedom left needs a stated ratio or split, not a guess.
Two reference states in one balance. Heats of formation and sensible enthalpies must refer to the same reference temperature and phase, or the energy balance carries a hidden latent heat.
"Steady" when it isn’t. A tank whose level changes has accumulation even if every flow is constant.
Sources: Felder and Rousseau, Elementary Principles of Chemical Processes, chapters on material and energy balances.