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 1250 kg/h · 15% sugarStream 2ṁ₂ = ? · 45% sugarProduct35% sugar

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

ṁ1 = 250 kg/hx1 = 15 %x2 = 45 %x3 = 35 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F600 kg/h · 7% NaOHProduct P45% 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 = 600 kg/hxF = 7 %xP = 45 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 180 °C120 °Cwater · 30 °Ct = ? °C

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

ṁh = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 180 → 120 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 150 °C90 °Cwater · 20 °C75 °C · ṁ = ?15% lost

Hot light oil 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 = 2.1 kJ/(kg·K)Th: 150 → 90 °Ct: 20 → 75 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
110 g of ArP = 500 kPaT = 40 °CV = ?

What volume V does 110 g of Ar occupy at 500 kPa and 40 °C? Treat it as an ideal gas.

m = 110 gM = 39.95 g/molP = 500 kPaT = 40 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 250 kPa, 10 L, 25 °C to 750 kPa and 70 °C. Find the new volume V2.

P1 = 250 kPaV1 = 10 LT1 = 25 °CP2 = 750 kPaT2 = 70 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

8 g of H2 and 48 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?

M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 g/mol
m = g
8.StoichiometryHard
m(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 6 g of Al reacts completely (the other reactants are in excess) and 20.76 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 20.76 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 500 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) = 250·(15 − 35)/(35 − 45) = 500 kg/h

Answer

ṁ2 = 500 kg/h
2.Mass balance

Answer: W = 506.7 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 = 600·7/45 = 93.33 kg/h

Total balance

W = F − P = 600 − 93.33 = 506.7 kg/h

Answer

W = 506.7 kg/h
3.Energy balance

Answer: tout = 82.54 °C

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

Heat released by the hot stream

Q = 3·2.44·(180 − 120) = 439.2 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 439.2/(2·4.18) = 82.54 °C

Answer

tout = 82.54 °C
4.Energy balance

Answer: ṁc = 0.6988 kg/s

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

Heat released

Qh = 1.5·2.1·(150 − 90) = 189 kW

Heat that reaches the water

Qc = (1 − 0.15)·189 = 160.7 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 160.7/(4.18·55) = 0.6988 kg/s

Answer

ṁc = 0.6988 kg/s
5.Ideal gas law

Answer: V = 14.34 L

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

Moles

n = 110/39.95 = 2.753 mol
T = 40 + 273.15 = 313.1 K

Volume

V = nRT/P = 2.753·8.314·313.1/500 = 14.34 L

Answer

V = 14.34 L
6.Ideal gas law

Answer: V2 = 3.836 L

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

Absolute temperatures

T1 = 298.1 K, T2 = 343.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(250/750)·(343.1/298.1) = 3.836 L

Answer

V2 = 3.836 L
7.Stoichiometry

Answer: m = 54.06 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(H2) = 8/2.016 = 3.968 mol
n(O2) = 48/32 = 1.5 mol

Limiting reactant

3.968/2 = 1.984, 1.5/1 = 1.5 → O2

Mass of product

n(H2O) = 1.5·2 = 3 mol
m = 3·18.02 = 54.06 g

Answer

m = 54.06 g
8.Stoichiometry

Answer: Y = 70.01 %

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(Al) = 6/26.98 = 0.2224 mol
m = 0.2224·2/2·133.33 = 29.65 g

Percent yield

Y = 20.76/29.65·100 = 70.01%

Answer

Y = 70.01%

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