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 1300 kg/h · 10% NaOHStream 2ṁ₂ = ? · 30% 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 = 300 kg/hx1 = 10 %x2 = 30 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F5000 kg/h · 11% sugarProduct P65% 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 = 5000 kg/hxF = 11 %xP = 65 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 140 °C60 °Cwater · 25 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 60 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 140 °C60 °Cwater · 15 °C35 °C · ṁ = ?10% lost

Hot glycol mix 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 = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 60 °Ct: 15 → 35 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
70 g of O₂P = 600 kPaT = 100 °CV = ?

What volume V does 70 g of O₂ occupy at 600 kPa and 100 °C? Treat it as an ideal gas.

m = 70 gM = 32 g/molP = 600 kPaT = 100 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 800 kPaT = 130 °Cρ = ?

Find the density ρ of CO₂ at 800 kPa and 130 °C, treating it as an ideal gas.

M = 44.01 g/molP = 800 kPaT = 130 °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)

24 g of Mg and 80 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(Fe)m(Fe)÷ M(Fe2​O3​)× 2/1× M(Fe)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 10 g of Fe2O3 reacts completely (the other reactants are in excess) and 4.34 g of Fe is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(Fe) = 55.85 g/molactual: 4.34 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 900 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) = 300·(10 − 25)/(25 − 30) = 900 kg/h

Answer

ṁ2 = 900 kg/h
2.Mass balance

Answer: W = 4154 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 = 5000·11/65 = 846.2 kg/h

Total balance

W = F − P = 5000 − 846.2 = 4154 kg/h

Answer

W = 4154 kg/h
3.Energy balance

Answer: tout = 58.49 °C

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

Heat released by the hot stream

Q = 2.5·2.1·(140 − 60) = 420 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 420/(3·4.18) = 58.49 °C

Answer

tout = 58.49 °C
4.Energy balance

Answer: ṁc = 4.651 kg/s

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

Heat released

Qh = 1.5·3.6·(140 − 60) = 432 kW

Heat that reaches the water

Qc = (1 − 0.1)·432 = 388.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 388.8/(4.18·20) = 4.651 kg/s

Answer

ṁc = 4.651 kg/s
5.Ideal gas law

Answer: V = 11.31 L

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

Moles

n = 70/32 = 2.188 mol
T = 100 + 273.15 = 373.1 K

Volume

V = nRT/P = 2.188·8.314·373.1/600 = 11.31 L

Answer

V = 11.31 L
6.Ideal gas law

Answer: ρ = 10.5 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 130 + 273.15 = 403.1 K

Density

ρ = 800·44.01/(8.314·403.1) = 10.5 kg/m³

Answer

ρ = 10.5 kg/m³
7.Stoichiometry

Answer: m = 39.79 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) = 24/24.31 = 0.9872 mol
n(O2) = 80/32 = 2.5 mol

Limiting reactant

0.9872/2 = 0.4936, 2.5/1 = 2.5 → Mg

Mass of product

n(MgO) = 0.4936·2 = 0.9872 mol
m = 0.9872·40.3 = 39.79 g

Answer

m = 39.79 g
8.Stoichiometry

Answer: Y = 62.05 %

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(Fe2O3) = 10/159.69 = 0.06262 mol
m = 0.06262·2/1·55.85 = 6.995 g

Percent yield

Y = 4.34/6.995·100 = 62.05%

Answer

Y = 62.05%

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