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 1200 kg/h · 20% NaOHStream 2ṁ₂ = ? · 30% NaOHProduct25% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 25% NaOH? Find ṁ2.

ṁ1 = 200 kg/hx1 = 20 %x2 = 30 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2400 kg/h · 12% saltProduct PP = ? · 50%Vapour W (pure water)0% salt

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

F = 2400 kg/hxF = 12 %xP = 50 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 130 °C60 °Cwater · 20 °Ct = ? °C

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

ṁh = 0.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 130 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 130 °C80 °Cwater · 10 °C85 °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: 130 → 80 °Ct: 10 → 85 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
330 g of HeV = 50 LT = 0 °CP = ?

A tank of 50 L contains 330 g of He at 0 °C. Treat it as an ideal gas and find the pressure P.

m = 330 gM = 4.003 g/molV = 50 LT = 0 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
Ar, M = 39.95 g/molP = 100 kPaT = 40 °Cρ = ?

Find the density ρ of Ar at 100 kPa and 40 °C, treating it as an ideal gas.

M = 39.95 g/molP = 100 kPaT = 40 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(AlCl3​)m(AlCl3​)÷ M(X)× 2/ν× M(AlCl3​)

8 g of Al and 8 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(H2​)n(H2​)n(H2​O)m(H2​O)÷ M(H2​)× 2/2× M(H2​O)

In the reaction 2H2 + O2 → 2H2O, 24 g of H2 reacts completely (the other reactants are in excess) and 167.3 g of H2O is collected. Find the percent yield Y.

M(H2) = 2.016 g/molM(H2O) = 18.02 g/molactual: 167.3 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 200 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) = 200·(20 − 25)/(25 − 30) = 200 kg/h

Answer

ṁ2 = 200 kg/h
2.Mass balance

Answer: P = 576 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 = 2400·12/50 = 576 kg/h

Answer

P = 576 kg/h
3.Energy balance

Answer: tout = 40.1 °C

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

Heat released by the hot stream

Q = 0.5·3.6·(130 − 60) = 126 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 126/(1.5·4.18) = 40.1 °C

Answer

tout = 40.1 °C
4.Energy balance

Answer: ṁc = 0.4962 kg/s

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

Heat released

Qh = 1.5·2.44·(130 − 80) = 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/(4.18·75) = 0.4962 kg/s

Answer

ṁc = 0.4962 kg/s
5.Ideal gas law

Answer: P = 3744 kPa

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

Moles

n = m/M = 330/4.003 = 82.44 mol
T = 0 + 273.15 = 273.1 K

Pressure

P = nRT/V = 82.44·8.314·273.1/0.05 = 3744 kPa

Answer

P = 3744 kPa
6.Ideal gas law

Answer: ρ = 1.534 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 40 + 273.15 = 313.1 K

Density

ρ = 100·39.95/(8.314·313.1) = 1.534 kg/m³

Answer

ρ = 1.534 kg/m³
7.Stoichiometry

Answer: m = 10.03 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) = 8/26.98 = 0.2965 mol
n(Cl2) = 8/70.9 = 0.1128 mol

Limiting reactant

0.2965/2 = 0.1483, 0.1128/3 = 0.03761 → Cl2

Mass of product

n(AlCl3) = 0.03761·2 = 0.07522 mol
m = 0.07522·133.33 = 10.03 g

Answer

m = 10.03 g
8.Stoichiometry

Answer: Y = 78 %

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(H2) = 24/2.016 = 11.9 mol
m = 11.9·2/2·18.02 = 214.5 g

Percent yield

Y = 167.3/214.5·100 = 78%

Answer

Y = 78%

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