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 1800 kg/h · 35% NaOHStream 2ṁ₂ = ? · 5% NaOHProduct30% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% NaOH? Find ṁ2.

ṁ1 = 800 kg/hx1 = 35 %x2 = 5 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2400 kg/h · 9% saltProduct P40% 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 = 2400 kg/hxF = 9 %xP = 40 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 120 °C80 °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 = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 120 → 80 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 150 °C90 °Cwater · 10 °C30 °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 = 2 kg/scp,h = 2.1 kJ/(kg·K)Th: 150 → 90 °Ct: 10 → 30 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
410 g of ArP = 550 kPaT = 85 °CV = ?

What volume V does 410 g of Ar occupy at 550 kPa and 85 °C? Treat it as an ideal gas.

m = 410 gM = 39.95 g/molP = 550 kPaT = 85 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of Ar goes from 150 kPa, 10 L, 10 °C to 150 kPa and 220 °C. Find the new volume V2.

P1 = 150 kPaV1 = 10 LT1 = 10 °CP2 = 150 kPaT2 = 220 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(AlCl3​)m(AlCl3​)÷ M(X)× 2/ν× M(AlCl3​)

30 g of Al and 10 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(N2​)n(N2​)n(NH3​)m(NH3​)÷ M(N2​)× 2/1× M(NH3​)

In the reaction N2 + 3H2 → 2NH3, 24 g of N2 reacts completely (the other reactants are in excess) and 24.8 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 24.8 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 160 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) = 800·(35 − 30)/(30 − 5) = 160 kg/h

Answer

ṁ2 = 160 kg/h
2.Mass balance

Answer: W = 1860 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 = 2400·9/40 = 540 kg/h

Total balance

W = F − P = 2400 − 540 = 1860 kg/h

Answer

W = 1860 kg/h
3.Energy balance

Answer: tout = 37.22 °C

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

Heat released by the hot stream

Q = 1·3.6·(120 − 80) = 144 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 144/(2·4.18) = 37.22 °C

Answer

tout = 37.22 °C
4.Energy balance

Answer: ṁc = 2.864 kg/s

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

Heat released

Qh = 2·2.1·(150 − 90) = 252 kW

Heat that reaches the water

Qc = (1 − 0.05)·252 = 239.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 239.4/(4.18·20) = 2.864 kg/s

Answer

ṁc = 2.864 kg/s
5.Ideal gas law

Answer: V = 55.56 L

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

Moles

n = 410/39.95 = 10.26 mol
T = 85 + 273.15 = 358.1 K

Volume

V = nRT/P = 10.26·8.314·358.1/550 = 55.56 L

Answer

V = 55.56 L
6.Ideal gas law

Answer: V2 = 17.42 L

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

Absolute temperatures

T1 = 283.1 K, T2 = 493.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(150/150)·(493.1/283.1) = 17.42 L

Answer

V2 = 17.42 L
7.Stoichiometry

Answer: m = 12.54 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) = 30/26.98 = 1.112 mol
n(Cl2) = 10/70.9 = 0.141 mol

Limiting reactant

1.112/2 = 0.556, 0.141/3 = 0.04701 → Cl2

Mass of product

n(AlCl3) = 0.04701·2 = 0.09403 mol
m = 0.09403·133.33 = 12.54 g

Answer

m = 12.54 g
8.Stoichiometry

Answer: Y = 85.01 %

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(N2) = 24/28.02 = 0.8565 mol
m = 0.8565·2/1·17.03 = 29.17 g

Percent yield

Y = 24.8/29.17·100 = 85.01%

Answer

Y = 85.01%

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