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 1650 kg/h · 45% NaOHStream 2ṁ₂ = ? · 15% NaOHProduct20% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 20% NaOH? Find ṁ2.

ṁ1 = 650 kg/hx1 = 45 %x2 = 15 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3600 kg/h · 14% saltProduct PP = ? · 65%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 = 3600 kg/hxF = 14 %xP = 65 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 130 °C100 °Cwater · 10 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 2.1 kJ/(kg·K)Th: 130 → 100 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 120 °C60 °Cwater · 25 °C75 °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: 120 → 60 °Ct: 25 → 75 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
300 g of N₂P = 300 kPaT = 45 °CV = ?

What volume V does 300 g of N₂ occupy at 300 kPa and 45 °C? Treat it as an ideal gas.

m = 300 gM = 28.01 g/molP = 300 kPaT = 45 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 550 kPaT = 160 °Cρ = ?

Find the density ρ of N₂ at 550 kPa and 160 °C, treating it as an ideal gas.

M = 28.01 g/molP = 550 kPaT = 160 °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)

8 g of Na and 20 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(CaCO3​)n(CaCO3​)n(CO2​)m(CO2​)÷ M(CaCO3​)× 1/1× M(CO2​)

In the reaction CaCO3 → CaO + CO2, 8 g of CaCO3 reacts completely (the other reactants are in excess) and 2.64 g of CO2 is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 2.64 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 3250 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) = 650·(45 − 20)/(20 − 15) = 3250 kg/h

Answer

ṁ2 = 3250 kg/h
2.Mass balance

Answer: P = 775.4 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 = 3600·14/65 = 775.4 kg/h

Answer

P = 775.4 kg/h
3.Energy balance

Answer: tout = 30.1 °C

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

Heat released by the hot stream

Q = 2·2.1·(130 − 100) = 126 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 126/(1.5·4.18) = 30.1 °C

Answer

tout = 30.1 °C
4.Energy balance

Answer: ṁc = 2.455 kg/s

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

Heat released

Qh = 2.5·3.6·(120 − 60) = 540 kW

Heat that reaches the water

Qc = (1 − 0.05)·540 = 513 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 513/(4.18·50) = 2.455 kg/s

Answer

ṁc = 2.455 kg/s
5.Ideal gas law

Answer: V = 94.43 L

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

Moles

n = 300/28.01 = 10.71 mol
T = 45 + 273.15 = 318.1 K

Volume

V = nRT/P = 10.71·8.314·318.1/300 = 94.43 L

Answer

V = 94.43 L
6.Ideal gas law

Answer: ρ = 4.278 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 160 + 273.15 = 433.1 K

Density

ρ = 550·28.01/(8.314·433.1) = 4.278 kg/m³

Answer

ρ = 4.278 kg/m³
7.Stoichiometry

Answer: m = 20.34 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) = 8/22.99 = 0.348 mol
n(Cl2) = 20/70.9 = 0.2821 mol

Limiting reactant

0.348/2 = 0.174, 0.2821/1 = 0.2821 → Na

Mass of product

n(NaCl) = 0.174·2 = 0.348 mol
m = 0.348·58.44 = 20.34 g

Answer

m = 20.34 g
8.Stoichiometry

Answer: Y = 75.05 %

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(CaCO3) = 8/100.09 = 0.07993 mol
m = 0.07993·1/1·44.01 = 3.518 g

Percent yield

Y = 2.64/3.518·100 = 75.05%

Answer

Y = 75.05%

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