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 1550 kg/h · 20% saltStream 2ṁ₂ = ? · 35% saltProduct30% salt

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

ṁ1 = 550 kg/hx1 = 20 %x2 = 35 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4700 kg/h · 10% saltProduct P50% 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 = 4700 kg/hxF = 10 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 170 °C110 °Cwater · 15 °Ct = ? °C

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

ṁh = 0.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 110 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 170 °C140 °Cwater · 15 °C65 °C · ṁ = ?5% lost

Hot glycol mix 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 = 2.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 170 → 140 °Ct: 15 → 65 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
120 g of O₂P = 100 kPaT = 5 °CV = ?

What volume V does 120 g of O₂ occupy at 100 kPa and 5 °C? Treat it as an ideal gas.

m = 120 gM = 32 g/molP = 100 kPaT = 5 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 450 kPaT = 50 °Cρ = ?

Find the density ρ of CO₂ at 450 kPa and 50 °C, treating it as an ideal gas.

M = 44.01 g/molP = 450 kPaT = 50 °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​)

30 g of Al and 20 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(Mg)n(Mg)n(MgO)m(MgO)÷ M(Mg)× 2/2× M(MgO)

In the reaction 2Mg + O2 → 2MgO, 10 g of Mg reacts completely (the other reactants are in excess) and 9.12 g of MgO is collected. Find the percent yield Y.

M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 9.12 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1100 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
salt: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 550·(20 − 30)/(30 − 35) = 1100 kg/h

Answer

ṁ2 = 1100 kg/h
2.Mass balance

Answer: W = 3760 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 = 4700·10/50 = 940 kg/h

Total balance

W = F − P = 4700 − 940 = 3760 kg/h

Answer

W = 3760 kg/h
3.Energy balance

Answer: tout = 23.76 °C

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

Heat released by the hot stream

Q = 0.5·2.44·(170 − 110) = 73.2 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 73.2/(2·4.18) = 23.76 °C

Answer

tout = 23.76 °C
4.Energy balance

Answer: ṁc = 1.227 kg/s

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

Heat released

Qh = 2.5·3.6·(170 − 140) = 270 kW

Heat that reaches the water

Qc = (1 − 0.05)·270 = 256.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 256.5/(4.18·50) = 1.227 kg/s

Answer

ṁc = 1.227 kg/s
5.Ideal gas law

Answer: V = 86.72 L

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

Moles

n = 120/32 = 3.75 mol
T = 5 + 273.15 = 278.1 K

Volume

V = nRT/P = 3.75·8.314·278.1/100 = 86.72 L

Answer

V = 86.72 L
6.Ideal gas law

Answer: ρ = 7.371 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 50 + 273.15 = 323.1 K

Density

ρ = 450·44.01/(8.314·323.1) = 7.371 kg/m³

Answer

ρ = 7.371 kg/m³
7.Stoichiometry

Answer: m = 25.07 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) = 20/70.9 = 0.2821 mol

Limiting reactant

1.112/2 = 0.556, 0.2821/3 = 0.09403 → Cl2

Mass of product

n(AlCl3) = 0.09403·2 = 0.1881 mol
m = 0.1881·133.33 = 25.07 g

Answer

m = 25.07 g
8.Stoichiometry

Answer: Y = 55.01 %

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(Mg) = 10/24.31 = 0.4114 mol
m = 0.4114·2/2·40.3 = 16.58 g

Percent yield

Y = 9.12/16.58·100 = 55.01%

Answer

Y = 55.01%

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