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% saltStream 2ṁ₂ = ? · 30% saltProduct20% salt

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

ṁ1 = 300 kg/hx1 = 10 %x2 = 30 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3500 kg/h · 13% NaOHProduct PP = ? · 40%Vapour W (pure water)0% NaOH

An evaporator concentrates a NaOH solution by boiling off pure water. Find the product flow P in kg/h.

F = 3500 kg/hxF = 13 %xP = 40 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 140 °C90 °Cwater · 15 °Ct = ? °C

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

ṁh = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 90 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 170 °C90 °Cwater · 25 °C55 °C · ṁ = ?5% lost

Hot light oil 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 kg/scp,h = 2.1 kJ/(kg·K)Th: 170 → 90 °Ct: 25 → 55 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
320 g of CH₄V = 30 LT = 20 °CP = ?

A tank of 30 L contains 320 g of CH₄ at 20 °C. Treat it as an ideal gas and find the pressure P.

m = 320 gM = 16.04 g/molV = 30 LT = 20 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 350 kPa, 5 L, 25 °C to 650 kPa and 170 °C. Find the new volume V2.

P1 = 350 kPaV1 = 5 LT1 = 25 °CP2 = 650 kPaT2 = 170 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

10 g of Mg and 64 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(C6​H12​O6​)n(C6​H12​O6​)n(CO2​)m(CO2​)÷ M(C6​H12​O6​)× 2/1× M(CO2​)

In the reaction C6H12O6 → 2C2H5OH + 2CO2, 4 g of C6H12O6 reacts completely (the other reactants are in excess) and 1.66 g of CO2 is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(CO2) = 44.01 g/molactual: 1.66 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) = 300·(10 − 20)/(20 − 30) = 300 kg/h

Answer

ṁ2 = 300 kg/h
2.Mass balance

Answer: P = 1138 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 = 3500·13/40 = 1138 kg/h

Answer

P = 1138 kg/h
3.Energy balance

Answer: tout = 27.56 °C

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

Heat released by the hot stream

Q = 1.5·2.1·(140 − 90) = 157.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 157.5/(3·4.18) = 27.56 °C

Answer

tout = 27.56 °C
4.Energy balance

Answer: ṁc = 2.545 kg/s

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

Heat released

Qh = 2·2.1·(170 − 90) = 336 kW

Heat that reaches the water

Qc = (1 − 0.05)·336 = 319.2 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 319.2/(4.18·30) = 2.545 kg/s

Answer

ṁc = 2.545 kg/s
5.Ideal gas law

Answer: P = 1621 kPa

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

Moles

n = m/M = 320/16.04 = 19.95 mol
T = 20 + 273.15 = 293.1 K

Pressure

P = nRT/V = 19.95·8.314·293.1/0.03 = 1621 kPa

Answer

P = 1621 kPa
6.Ideal gas law

Answer: V2 = 4.002 L

Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.

Absolute temperatures

T1 = 298.1 K, T2 = 443.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(350/650)·(443.1/298.1) = 4.002 L

Answer

V2 = 4.002 L
7.Stoichiometry

Answer: m = 16.58 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) = 10/24.31 = 0.4114 mol
n(O2) = 64/32 = 2 mol

Limiting reactant

0.4114/2 = 0.2057, 2/1 = 2 → Mg

Mass of product

n(MgO) = 0.2057·2 = 0.4114 mol
m = 0.4114·40.3 = 16.58 g

Answer

m = 16.58 g
8.Stoichiometry

Answer: Y = 84.94 %

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) = 4/180.16 = 0.0222 mol
m = 0.0222·2/1·44.01 = 1.954 g

Percent yield

Y = 1.66/1.954·100 = 84.94%

Answer

Y = 84.94%

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