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 1750 kg/h · 50% NaOHStream 2ṁ₂ = ? · 20% NaOHProduct45% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 45% NaOH? Find ṁ2.

ṁ1 = 750 kg/hx1 = 50 %x2 = 20 %x3 = 45 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4900 kg/h · 19% saltProduct PP = ? · 40%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 = 4900 kg/hxF = 19 %xP = 40 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 120 °C60 °Cwater · 20 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 160 °C100 °Cwater · 10 °C40 °C · ṁ = ?5% lost

Hot light oil 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 = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 160 → 100 °Ct: 10 → 40 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
410 g of N₂P = 550 kPaT = 50 °CV = ?

What volume V does 410 g of N₂ occupy at 550 kPa and 50 °C? Treat it as an ideal gas.

m = 410 gM = 28.01 g/molP = 550 kPaT = 50 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of CO₂ goes from 200 kPa, 20 L, 35 °C to 400 kPa and 240 °C. Find the new volume V2.

P1 = 200 kPaV1 = 20 LT1 = 35 °CP2 = 400 kPaT2 = 240 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NaCl)m(NaCl)÷ M(X)× 2/ν× M(NaCl)

16 g of Na and 32 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?

M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 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, 10 g of CH4 reacts completely (the other reactants are in excess) and 17.53 g of H2O is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(H2O) = 18.02 g/molactual: 17.53 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 150 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) = 750·(50 − 45)/(45 − 20) = 150 kg/h

Answer

ṁ2 = 150 kg/h
2.Mass balance

Answer: P = 2328 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 = 4900·19/40 = 2328 kg/h

Answer

P = 2328 kg/h
3.Energy balance

Answer: tout = 78.37 °C

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

Heat released by the hot stream

Q = 2.5·2.44·(120 − 60) = 366 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 366/(1.5·4.18) = 78.37 °C

Answer

tout = 78.37 °C
4.Energy balance

Answer: ṁc = 0.4773 kg/s

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

Heat released

Qh = 0.5·2.1·(160 − 100) = 63 kW

Heat that reaches the water

Qc = (1 − 0.05)·63 = 59.85 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 59.85/(4.18·30) = 0.4773 kg/s

Answer

ṁc = 0.4773 kg/s
5.Ideal gas law

Answer: V = 71.5 L

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

Moles

n = 410/28.01 = 14.64 mol
T = 50 + 273.15 = 323.1 K

Volume

V = nRT/P = 14.64·8.314·323.1/550 = 71.5 L

Answer

V = 71.5 L
6.Ideal gas law

Answer: V2 = 16.65 L

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

Absolute temperatures

T1 = 308.1 K, T2 = 513.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(200/400)·(513.1/308.1) = 16.65 L

Answer

V2 = 16.65 L
7.Stoichiometry

Answer: m = 40.67 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(Na) = 16/22.99 = 0.696 mol
n(Cl2) = 32/70.9 = 0.4513 mol

Limiting reactant

0.696/2 = 0.348, 0.4513/1 = 0.4513 → Na

Mass of product

n(NaCl) = 0.348·2 = 0.696 mol
m = 0.696·58.44 = 40.67 g

Answer

m = 40.67 g
8.Stoichiometry

Answer: Y = 78.02 %

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) = 10/16.04 = 0.6234 mol
m = 0.6234·2/1·18.02 = 22.47 g

Percent yield

Y = 17.53/22.47·100 = 78.02%

Answer

Y = 78.02%

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