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 1850 kg/h · 40% NaOHStream 2ṁ₂ = ? · 10% NaOHProduct25% NaOH

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

ṁ1 = 850 kg/hx1 = 40 %x2 = 10 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F500 kg/h · 14% sugarProduct P60% sugarVapour W (pure water)W = ? kg/h

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

F = 500 kg/hxF = 14 %xP = 60 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 180 °C130 °Cwater · 20 °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: 180 → 130 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 170 °C80 °Cwater · 15 °C75 °C · ṁ = ?10% lost

Hot ethanol heats water in an exchanger that loses 10% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 80 °Ct: 15 → 75 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
190 g of HeV = 20 LT = 20 °CP = ?

A tank of 20 L contains 190 g of He at 20 °C. Treat it as an ideal gas and find the pressure P.

m = 190 gM = 4.003 g/molV = 20 LT = 20 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 200 kPaT = 170 °Cρ = ?

Find the density ρ of CO₂ at 200 kPa and 170 °C, treating it as an ideal gas.

M = 44.01 g/molP = 200 kPaT = 170 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 3/ν× M(CO2​)

30 g of Fe2O3 and 48 g of CO react according to Fe2O3 + 3CO → 2Fe + 3CO2. What mass m of CO2 can form?

M(Fe2O3) = 159.69M(CO) = 28.01M(CO2) = 44.01 g/mol
m = g
8.StoichiometryHard
m(C6​H12​O6​)n(C6​H12​O6​)n(C2​H5​OH)m(C2​H5​OH)÷ M(C6​H12​O6​)× 2/1× M(C2​H5​OH)

In the reaction C6H12O6 → 2C2H5OH + 2CO2, 24 g of C6H12O6 reacts completely (the other reactants are in excess) and 10.43 g of C2H5OH is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/molactual: 10.43 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 850 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) = 850·(40 − 25)/(25 − 10) = 850 kg/h

Answer

ṁ2 = 850 kg/h
2.Mass balance

Answer: W = 383.3 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 = 500·14/60 = 116.7 kg/h

Total balance

W = F − P = 500 − 116.7 = 383.3 kg/h

Answer

W = 383.3 kg/h
3.Energy balance

Answer: tout = 29.73 °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·(180 − 130) = 61 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 61/(1.5·4.18) = 29.73 °C

Answer

tout = 29.73 °C
4.Energy balance

Answer: ṁc = 2.364 kg/s

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

Heat released

Qh = 3·2.44·(170 − 80) = 658.8 kW

Heat that reaches the water

Qc = (1 − 0.1)·658.8 = 592.9 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 592.9/(4.18·60) = 2.364 kg/s

Answer

ṁc = 2.364 kg/s
5.Ideal gas law

Answer: P = 5784 kPa

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

Moles

n = m/M = 190/4.003 = 47.46 mol
T = 20 + 273.15 = 293.1 K

Pressure

P = nRT/V = 47.46·8.314·293.1/0.02 = 5784 kPa

Answer

P = 5784 kPa
6.Ideal gas law

Answer: ρ = 2.389 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 170 + 273.15 = 443.1 K

Density

ρ = 200·44.01/(8.314·443.1) = 2.389 kg/m³

Answer

ρ = 2.389 kg/m³
7.Stoichiometry

Answer: m = 24.8 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(Fe2O3) = 30/159.69 = 0.1879 mol
n(CO) = 48/28.01 = 1.714 mol

Limiting reactant

0.1879/1 = 0.1879, 1.714/3 = 0.5712 → Fe2O3

Mass of product

n(CO2) = 0.1879·3 = 0.5636 mol
m = 0.5636·44.01 = 24.8 g

Answer

m = 24.8 g
8.Stoichiometry

Answer: Y = 84.97 %

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(C6H12O6) = 24/180.16 = 0.1332 mol
m = 0.1332·2/1·46.07 = 12.27 g

Percent yield

Y = 10.43/12.27·100 = 84.97%

Answer

Y = 84.97%

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