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.

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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 11000 kg/h · 15% sugarStream 2ṁ₂ = ? · 35% sugarProduct20% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 20% sugar? Find ṁ2.

ṁ1 = 1000 kg/hx1 = 15 %x2 = 35 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3500 kg/h · 9% NaOHProduct PP = ? · 30%Vapour W (pure water)0% NaOH

An evaporator concentrates a NaOH solution by boiling off pure water. Find the product flow P in kg/h.

F = 3500 kg/hxF = 9 %xP = 30 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 120 °C90 °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: 120 → 90 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 120 °C70 °Cwater · 25 °C90 °C · ṁ = ?15% lost

Hot ethanol heats water in an exchanger that loses 15% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 1.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 70 °Ct: 25 → 90 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
480 g of CH₄P = 500 kPaT = 10 °CV = ?

What volume V does 480 g of CH₄ occupy at 500 kPa and 10 °C? Treat it as an ideal gas.

m = 480 gM = 16.04 g/molP = 500 kPaT = 10 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 250 kPa, 5 L, 30 °C to 400 kPa and 90 °C. Find the new volume V2.

P1 = 250 kPaV1 = 5 LT1 = 30 °CP2 = 400 kPaT2 = 90 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(AlCl3​)m(AlCl3​)÷ M(X)× 2/ν× M(AlCl3​)

24 g of Al and 20 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(H2​O)m(H2​O)÷ M(CH4​)× 2/1× M(H2​O)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 8 g of CH4 reacts completely (the other reactants are in excess) and 13.48 g of H2O is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(H2O) = 18.02 g/molactual: 13.48 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 333.3 kg/h

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

Balances

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

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 1000·(15 − 20)/(20 − 35) = 333.3 kg/h

Answer

ṁ2 = 333.3 kg/h
2.Mass balance

Answer: P = 1050 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 = 3500·9/30 = 1050 kg/h

Answer

P = 1050 kg/h
3.Energy balance

Answer: tout = 45.07 °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·(120 − 90) = 189 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 189/(3·4.18) = 45.07 °C

Answer

tout = 45.07 °C
4.Energy balance

Answer: ṁc = 0.5725 kg/s

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

Heat released

Qh = 1.5·2.44·(120 − 70) = 183 kW

Heat that reaches the water

Qc = (1 − 0.15)·183 = 155.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 155.5/(4.18·65) = 0.5725 kg/s

Answer

ṁc = 0.5725 kg/s
5.Ideal gas law

Answer: V = 140.9 L

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

Moles

n = 480/16.04 = 29.93 mol
T = 10 + 273.15 = 283.1 K

Volume

V = nRT/P = 29.93·8.314·283.1/500 = 140.9 L

Answer

V = 140.9 L
6.Ideal gas law

Answer: V2 = 3.744 L

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

Absolute temperatures

T1 = 303.1 K, T2 = 363.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(250/400)·(363.1/303.1) = 3.744 L

Answer

V2 = 3.744 L
7.Stoichiometry

Answer: m = 25.07 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) = 24/26.98 = 0.8895 mol
n(Cl2) = 20/70.9 = 0.2821 mol

Limiting reactant

0.8895/2 = 0.4448, 0.2821/3 = 0.09403 → Cl2

Mass of product

n(AlCl3) = 0.09403·2 = 0.1881 mol
m = 0.1881·133.33 = 25.07 g

Answer

m = 25.07 g
8.Stoichiometry

Answer: Y = 74.99 %

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) = 8/16.04 = 0.4988 mol
m = 0.4988·2/1·18.02 = 17.98 g

Percent yield

Y = 13.48/17.98·100 = 74.99%

Answer

Y = 74.99%

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