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 11000 kg/h · 20% saltStream 2ṁ₂ = ? · 40% 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 = 1000 kg/hx1 = 20 %x2 = 40 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2400 kg/h · 16% saltProduct PP = ? · 45%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 = 2400 kg/hxF = 16 %xP = 45 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 140 °C100 °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 = 1 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 100 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 150 °C60 °Cwater · 25 °C80 °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 = 1.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 150 → 60 °Ct: 25 → 80 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
220 g of ArV = 20 LT = 65 °CP = ?

A tank of 20 L contains 220 g of Ar at 65 °C. Treat it as an ideal gas and find the pressure P.

m = 220 gM = 39.95 g/molV = 20 LT = 65 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 600 kPaT = 0 °Cρ = ?

Find the density ρ of CO₂ at 600 kPa and 0 °C, treating it as an ideal gas.

M = 44.01 g/molP = 600 kPaT = 0 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

16 g of Mg and 40 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?

M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 g/mol
m = g
8.StoichiometryHard
m(Fe2​O3​)n(Fe2​O3​)n(CO2​)m(CO2​)÷ M(Fe2​O3​)× 3/1× M(CO2​)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 20 g of Fe2O3 reacts completely (the other reactants are in excess) and 14.06 g of CO2 is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/molactual: 14.06 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 333.3 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) = 1000·(20 − 25)/(25 − 40) = 333.3 kg/h

Answer

ṁ2 = 333.3 kg/h
2.Mass balance

Answer: P = 853.3 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 = 2400·16/45 = 853.3 kg/h

Answer

P = 853.3 kg/h
3.Energy balance

Answer: tout = 27.78 °C

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

Heat released by the hot stream

Q = 1·2.44·(140 − 100) = 97.6 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 97.6/(3·4.18) = 27.78 °C

Answer

tout = 27.78 °C
4.Energy balance

Answer: ṁc = 1.361 kg/s

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

Heat released

Qh = 1.5·2.44·(150 − 60) = 329.4 kW

Heat that reaches the water

Qc = (1 − 0.05)·329.4 = 312.9 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 312.9/(4.18·55) = 1.361 kg/s

Answer

ṁc = 1.361 kg/s
5.Ideal gas law

Answer: P = 774.1 kPa

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

Moles

n = m/M = 220/39.95 = 5.507 mol
T = 65 + 273.15 = 338.1 K

Pressure

P = nRT/V = 5.507·8.314·338.1/0.02 = 774.1 kPa

Answer

P = 774.1 kPa
6.Ideal gas law

Answer: ρ = 11.63 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 0 + 273.15 = 273.1 K

Density

ρ = 600·44.01/(8.314·273.1) = 11.63 kg/m³

Answer

ρ = 11.63 kg/m³
7.Stoichiometry

Answer: m = 26.52 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(Mg) = 16/24.31 = 0.6582 mol
n(O2) = 40/32 = 1.25 mol

Limiting reactant

0.6582/2 = 0.3291, 1.25/1 = 1.25 → Mg

Mass of product

n(MgO) = 0.3291·2 = 0.6582 mol
m = 0.6582·40.3 = 26.52 g

Answer

m = 26.52 g
8.Stoichiometry

Answer: Y = 85.03 %

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(Fe2O3) = 20/159.69 = 0.1252 mol
m = 0.1252·3/1·44.01 = 16.54 g

Percent yield

Y = 14.06/16.54·100 = 85.03%

Answer

Y = 85.03%

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