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 1450 kg/h · 50% NaOHStream 2ṁ₂ = ? · 10% NaOHProduct15% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 15% NaOH? Find ṁ2.

ṁ1 = 450 kg/hx1 = 50 %x2 = 10 %x3 = 15 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4100 kg/h · 13% saltProduct P30% saltVapour W (pure water)W = ? kg/h

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

F = 4100 kg/hxF = 13 %xP = 30 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 120 °C80 °Cwater · 25 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 80 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 130 °C100 °Cwater · 25 °C45 °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: 130 → 100 °Ct: 25 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
230 g of N₂P = 500 kPaT = 25 °CV = ?

What volume V does 230 g of N₂ occupy at 500 kPa and 25 °C? Treat it as an ideal gas.

m = 230 gM = 28.01 g/molP = 500 kPaT = 25 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of O₂ goes from 100 kPa, 5 L, 10 °C to 700 kPa and 180 °C. Find the new volume V2.

P1 = 100 kPaV1 = 5 LT1 = 10 °CP2 = 700 kPaT2 = 180 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 1/ν× M(CO2​)

10 g of CH4 and 64 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of CO2 can form?

M(CH4) = 16.04M(O2) = 32M(CO2) = 44.01 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, 8 g of H2 reacts completely (the other reactants are in excess) and 60.78 g of H2O is collected. Find the percent yield Y.

M(H2) = 2.016 g/molM(H2O) = 18.02 g/molactual: 60.78 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 3150 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) = 450·(50 − 15)/(15 − 10) = 3150 kg/h

Answer

ṁ2 = 3150 kg/h
2.Mass balance

Answer: W = 2323 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 = 4100·13/30 = 1777 kg/h

Total balance

W = F − P = 4100 − 1777 = 2323 kg/h

Answer

W = 2323 kg/h
3.Energy balance

Answer: tout = 48.35 °C

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

Heat released by the hot stream

Q = 2·2.44·(120 − 80) = 195.2 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 195.2/(2·4.18) = 48.35 °C

Answer

tout = 48.35 °C
4.Energy balance

Answer: ṁc = 0.8318 kg/s

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

Heat released

Qh = 1·2.44·(130 − 100) = 73.2 kW

Heat that reaches the water

Qc = (1 − 0.05)·73.2 = 69.54 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 69.54/(4.18·20) = 0.8318 kg/s

Answer

ṁc = 0.8318 kg/s
5.Ideal gas law

Answer: V = 40.71 L

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

Moles

n = 230/28.01 = 8.211 mol
T = 25 + 273.15 = 298.1 K

Volume

V = nRT/P = 8.211·8.314·298.1/500 = 40.71 L

Answer

V = 40.71 L
6.Ideal gas law

Answer: V2 = 1.143 L

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

Absolute temperatures

T1 = 283.1 K, T2 = 453.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(100/700)·(453.1/283.1) = 1.143 L

Answer

V2 = 1.143 L
7.Stoichiometry

Answer: m = 27.44 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(CH4) = 10/16.04 = 0.6234 mol
n(O2) = 64/32 = 2 mol

Limiting reactant

0.6234/1 = 0.6234, 2/2 = 1 → CH4

Mass of product

n(CO2) = 0.6234·1 = 0.6234 mol
m = 0.6234·44.01 = 27.44 g

Answer

m = 27.44 g
8.Stoichiometry

Answer: Y = 85 %

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) = 8/2.016 = 3.968 mol
m = 3.968·2/2·18.02 = 71.51 g

Percent yield

Y = 60.78/71.51·100 = 85%

Answer

Y = 85%

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