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 · 15% saltStream 2ṁ₂ = ? · 40% 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 = 850 kg/hx1 = 15 %x2 = 40 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4100 kg/h · 5% NaOHProduct P55% NaOHVapour W (pure water)W = ? kg/h

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

F = 4100 kg/hxF = 5 %xP = 55 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 160 °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: 160 → 130 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 180 °C70 °Cwater · 20 °C70 °C · ṁ = ?15% lost

Hot glycol mix heats water in an exchanger that loses 15% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 2.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 180 → 70 °Ct: 20 → 70 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
130 g of O₂P = 250 kPaT = 30 °CV = ?

What volume V does 130 g of O₂ occupy at 250 kPa and 30 °C? Treat it as an ideal gas.

m = 130 gM = 32 g/molP = 250 kPaT = 30 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of O₂ goes from 200 kPa, 10 L, 40 °C to 400 kPa and 110 °C. Find the new volume V2.

P1 = 200 kPaV1 = 10 LT1 = 40 °CP2 = 400 kPaT2 = 110 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

30 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, 24 g of Fe2O3 reacts completely (the other reactants are in excess) and 12.3 g of CO2 is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/molactual: 12.3 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 212.5 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) = 850·(15 − 20)/(20 − 40) = 212.5 kg/h

Answer

ṁ2 = 212.5 kg/h
2.Mass balance

Answer: W = 3727 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 = 4100·5/55 = 372.7 kg/h

Total balance

W = F − P = 4100 − 372.7 = 3727 kg/h

Answer

W = 3727 kg/h
3.Energy balance

Answer: tout = 28.76 °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·(160 − 130) = 36.6 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 36.6/(1·4.18) = 28.76 °C

Answer

tout = 28.76 °C
4.Energy balance

Answer: ṁc = 4.026 kg/s

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

Heat released

Qh = 2.5·3.6·(180 − 70) = 990 kW

Heat that reaches the water

Qc = (1 − 0.15)·990 = 841.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 841.5/(4.18·50) = 4.026 kg/s

Answer

ṁc = 4.026 kg/s
5.Ideal gas law

Answer: V = 40.96 L

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

Moles

n = 130/32 = 4.063 mol
T = 30 + 273.15 = 303.1 K

Volume

V = nRT/P = 4.063·8.314·303.1/250 = 40.96 L

Answer

V = 40.96 L
6.Ideal gas law

Answer: V2 = 6.118 L

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

Absolute temperatures

T1 = 313.1 K, T2 = 383.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(200/400)·(383.1/313.1) = 6.118 L

Answer

V2 = 6.118 L
7.Stoichiometry

Answer: m = 49.73 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) = 30/24.31 = 1.234 mol
n(O2) = 40/32 = 1.25 mol

Limiting reactant

1.234/2 = 0.617, 1.25/1 = 1.25 → Mg

Mass of product

n(MgO) = 0.617·2 = 1.234 mol
m = 1.234·40.3 = 49.73 g

Answer

m = 49.73 g
8.Stoichiometry

Answer: Y = 61.99 %

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) = 24/159.69 = 0.1503 mol
m = 0.1503·3/1·44.01 = 19.84 g

Percent yield

Y = 12.3/19.84·100 = 61.99%

Answer

Y = 61.99%

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