Material and Energy Balances: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

Mass balance

Steady state
in = out
Total balance
ṁ1 + ṁ2 = ṁ3
Component balance
ṁ1x1 + ṁ2x2 = ṁ3x3
Evaporator
F·xF = P·xP, W = F − P

Energy balance

Sensible heat
Q = ṁ·cp·ΔT
Latent heat
Q = ṁ·hfg
Mixing
ΣṁcpTin = ΣṁcpTout
Water
cp ≈ 4.18 kJ/(kg·K)

Ideal gas law

Ideal gas law
pV = nRT
Gas constant
R = 8.314 J/(mol·K)
Combined gas law
p1V1/T1 = p2V2/T2
Moles
n = m/M
Temperature
T[K] = T[°C] + 273.15

Stoichiometry

Moles
n = m/M
Concentration
c = n/V
Mole ratio
use the coefficients of the balanced equation
Limiting reactant
the one that runs out first
Percent yield
actual / theoretical × 100%

Method

  1. Write and balance the chemical equation.
  2. Convert the given mass to moles: n = m/M.
  3. Use the coefficients as the mole ratio to find the moles of the substance you want.
  4. Convert back to mass with m = n·M. With two reactants, find the limiting reactant first.

2Practice problems

1.Mass balanceMedium
MixerStream 1950 kg/h · 40% NaOHStream 2ṁ₂ = ? · 15% NaOHProduct35% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 35% NaOH? Find ṁ2.

ṁ1 = 950 kg/hx1 = 40 %x2 = 15 %x3 = 35 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3200 kg/h · 19% NaOHProduct P40% NaOHVapour W (pure water)W = ? kg/h

An evaporator concentrates a NaOH solution by boiling off pure water. Find the water evaporated W in kg/h.

F = 3200 kg/hxF = 19 %xP = 40 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 170 °C130 °Cwater · 30 °Ct = ? °C

Hot light oil cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature tout.

ṁh = 3 kg/scp,h = 2.1 kJ/(kg·K)Th: 170 → 130 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 160 °C100 °Cwater · 10 °C60 °C · ṁ = ?5% lost

Hot ethanol heats water in an exchanger that loses 5% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 1 kg/scp,h = 2.44 kJ/(kg·K)Th: 160 → 100 °Ct: 10 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
360 g of CO₂P = 800 kPaT = 70 °CV = ?

What volume V does 360 g of CO₂ occupy at 800 kPa and 70 °C? Treat it as an ideal gas.

m = 360 gM = 44.01 g/molP = 800 kPaT = 70 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 400 kPa, 10 L, 25 °C to 800 kPa and 250 °C. Find the new volume V2.

P1 = 400 kPaV1 = 10 LT1 = 25 °CP2 = 800 kPaT2 = 250 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(AlCl3​)m(AlCl3​)÷ M(X)× 2/ν× M(AlCl3​)

6 g of Al and 24 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?

M(Al) = 26.98M(Cl2) = 70.9M(AlCl3) = 133.33 g/mol
m = g
8.StoichiometryHard
m(CH4​)n(CH4​)n(CO2​)m(CO2​)÷ M(CH4​)× 1/1× M(CO2​)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 6 g of CH4 reacts completely (the other reactants are in excess) and 10.21 g of CO2 is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(CO2) = 44.01 g/molactual: 10.21 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 237.5 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
NaOH: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 950·(40 − 35)/(35 − 15) = 237.5 kg/h

Answer

ṁ2 = 237.5 kg/h
2.Mass balance

Answer: W = 1680 kg/h

Tie component: a component that passes through unchanged in one stream (here the solute) links the flows directly. Then the total balance gives the rest.

Solute balance

F·xF = P·xP → P = 3200·19/40 = 1520 kg/h

Total balance

W = F − P = 3200 − 1520 = 1680 kg/h

Answer

W = 1680 kg/h
3.Energy balance

Answer: tout = 70.19 °C

Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).

Heat released by the hot stream

Q = 3·2.1·(170 − 130) = 252 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 252/(1.5·4.18) = 70.19 °C

Answer

tout = 70.19 °C
4.Energy balance

Answer: ṁc = 0.6655 kg/s

Energy balance with losses: heat released = heat absorbed + heat lost.

Heat released

Qh = 1·2.44·(160 − 100) = 146.4 kW

Heat that reaches the water

Qc = (1 − 0.05)·146.4 = 139.1 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 139.1/(4.18·50) = 0.6655 kg/s

Answer

ṁc = 0.6655 kg/s
5.Ideal gas law

Answer: V = 29.17 L

Ideal gas law: PV = nRT, with n = m/M.

Moles

n = 360/44.01 = 8.18 mol
T = 70 + 273.15 = 343.1 K

Volume

V = nRT/P = 8.18·8.314·343.1/800 = 29.17 L

Answer

V = 29.17 L
6.Ideal gas law

Answer: V2 = 8.773 L

Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.

Absolute temperatures

T1 = 298.1 K, T2 = 523.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(400/800)·(523.1/298.1) = 8.773 L

Answer

V2 = 8.773 L
7.Stoichiometry

Answer: m = 29.65 g

Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.

Moles of each reactant

n(Al) = 6/26.98 = 0.2224 mol
n(Cl2) = 24/70.9 = 0.3385 mol

Limiting reactant

0.2224/2 = 0.1112, 0.3385/3 = 0.1128 → Al

Mass of product

n(AlCl3) = 0.1112·2 = 0.2224 mol
m = 0.2224·133.33 = 29.65 g

Answer

m = 29.65 g
8.Stoichiometry

Answer: Y = 62.02 %

Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.

Theoretical yield

n(CH4) = 6/16.04 = 0.3741 mol
m = 0.3741·1/1·44.01 = 16.46 g

Percent yield

Y = 10.21/16.46·100 = 62.02%

Answer

Y = 62.02%

ExamProblems · examproblems.com/courses/material-and-energy-balances