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 1750 kg/h · 5% NaOHStream 2ṁ₂ = ? · 35% NaOHProduct30% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% NaOH? Find ṁ2.

ṁ1 = 750 kg/hx1 = 5 %x2 = 35 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3200 kg/h · 8% saltProduct PP = ? · 30%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 = 3200 kg/hxF = 8 %xP = 30 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 180 °C80 °Cwater · 25 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 2.44 kJ/(kg·K)Th: 180 → 80 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 180 °C110 °Cwater · 15 °C40 °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 kg/scp,h = 3.6 kJ/(kg·K)Th: 180 → 110 °Ct: 15 → 40 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
380 g of CH₄V = 30 LT = 35 °CP = ?

A tank of 30 L contains 380 g of CH₄ at 35 °C. Treat it as an ideal gas and find the pressure P.

m = 380 gM = 16.04 g/molV = 30 LT = 35 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of N₂ goes from 100 kPa, 10 L, 30 °C to 350 kPa and 110 °C. Find the new volume V2.

P1 = 100 kPaV1 = 10 LT1 = 30 °CP2 = 350 kPaT2 = 110 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

24 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(C2​H5​OH)m(C2​H5​OH)÷ M(C6​H12​O6​)× 2/1× M(C2​H5​OH)

In the reaction C6H12O6 → 2C2H5OH + 2CO2, 6 g of C6H12O6 reacts completely (the other reactants are in excess) and 1.9 g of C2H5OH is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/molactual: 1.9 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 3750 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) = 750·(5 − 30)/(30 − 35) = 3750 kg/h

Answer

ṁ2 = 3750 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 = 3200·8/30 = 853.3 kg/h

Answer

P = 853.3 kg/h
3.Energy balance

Answer: tout = 63.92 °C

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

Heat released by the hot stream

Q = 2·2.44·(180 − 80) = 488 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 488/(3·4.18) = 63.92 °C

Answer

tout = 63.92 °C
4.Energy balance

Answer: ṁc = 2.17 kg/s

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

Heat released

Qh = 1·3.6·(180 − 110) = 252 kW

Heat that reaches the water

Qc = (1 − 0.1)·252 = 226.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 226.8/(4.18·25) = 2.17 kg/s

Answer

ṁc = 2.17 kg/s
5.Ideal gas law

Answer: P = 2023 kPa

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

Moles

n = m/M = 380/16.04 = 23.69 mol
T = 35 + 273.15 = 308.1 K

Pressure

P = nRT/V = 23.69·8.314·308.1/0.03 = 2023 kPa

Answer

P = 2023 kPa
6.Ideal gas law

Answer: V2 = 3.611 L

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

Absolute temperatures

T1 = 303.1 K, T2 = 383.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(100/350)·(383.1/303.1) = 3.611 L

Answer

V2 = 3.611 L
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) = 64/32 = 2 mol

Limiting reactant

0.9872/2 = 0.4936, 2/1 = 2 → 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 = 61.92 %

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) = 6/180.16 = 0.0333 mol
m = 0.0333·2/1·46.07 = 3.069 g

Percent yield

Y = 1.9/3.069·100 = 61.92%

Answer

Y = 61.92%

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