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 1200 kg/h · 10% saltStream 2ṁ₂ = ? · 35% saltProduct25% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 25% salt? Find ṁ2.

ṁ1 = 200 kg/hx1 = 10 %x2 = 35 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1400 kg/h · 11% 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 = 1400 kg/hxF = 11 %xP = 45 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 140 °C80 °Cwater · 15 °Ct = ? °C

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

ṁh = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 80 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 160 °C130 °Cwater · 15 °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 = 2.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 160 → 130 °Ct: 15 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
390 g of N₂P = 550 kPaT = 95 °CV = ?

What volume V does 390 g of N₂ occupy at 550 kPa and 95 °C? Treat it as an ideal gas.

m = 390 gM = 28.01 g/molP = 550 kPaT = 95 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CH₄, M = 16.04 g/molP = 700 kPaT = 200 °Cρ = ?

Find the density ρ of CH₄ at 700 kPa and 200 °C, treating it as an ideal gas.

M = 16.04 g/molP = 700 kPaT = 200 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(NaCl)m(NaCl)÷ M(X)× 2/ν× M(NaCl)

10 g of Na and 16 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(N2​)n(N2​)n(NH3​)m(NH3​)÷ M(N2​)× 2/1× M(NH3​)

In the reaction N2 + 3H2 → 2NH3, 8 g of N2 reacts completely (the other reactants are in excess) and 6.03 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 6.03 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 300 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) = 200·(10 − 25)/(25 − 35) = 300 kg/h

Answer

ṁ2 = 300 kg/h
2.Mass balance

Answer: W = 1058 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 = 1400·11/45 = 342.2 kg/h

Total balance

W = F − P = 1400 − 342.2 = 1058 kg/h

Answer

W = 1058 kg/h
3.Energy balance

Answer: tout = 40.84 °C

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

Heat released by the hot stream

Q = 1.5·3.6·(140 − 80) = 324 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 324/(3·4.18) = 40.84 °C

Answer

tout = 40.84 °C
4.Energy balance

Answer: ṁc = 1.386 kg/s

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

Heat released

Qh = 2.5·2.44·(160 − 130) = 183 kW

Heat that reaches the water

Qc = (1 − 0.05)·183 = 173.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 173.8/(4.18·30) = 1.386 kg/s

Answer

ṁc = 1.386 kg/s
5.Ideal gas law

Answer: V = 77.49 L

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

Moles

n = 390/28.01 = 13.92 mol
T = 95 + 273.15 = 368.1 K

Volume

V = nRT/P = 13.92·8.314·368.1/550 = 77.49 L

Answer

V = 77.49 L
6.Ideal gas law

Answer: ρ = 2.854 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 200 + 273.15 = 473.1 K

Density

ρ = 700·16.04/(8.314·473.1) = 2.854 kg/m³

Answer

ρ = 2.854 kg/m³
7.Stoichiometry

Answer: m = 25.42 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) = 10/22.99 = 0.435 mol
n(Cl2) = 16/70.9 = 0.2257 mol

Limiting reactant

0.435/2 = 0.2175, 0.2257/1 = 0.2257 → Na

Mass of product

n(NaCl) = 0.2175·2 = 0.435 mol
m = 0.435·58.44 = 25.42 g

Answer

m = 25.42 g
8.Stoichiometry

Answer: Y = 62.01 %

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) = 8/28.02 = 0.2855 mol
m = 0.2855·2/1·17.03 = 9.724 g

Percent yield

Y = 6.03/9.724·100 = 62.01%

Answer

Y = 62.01%

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