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 1550 kg/h · 15% saltStream 2ṁ₂ = ? · 35% 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 = 550 kg/hx1 = 15 %x2 = 35 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3300 kg/h · 11% saltProduct PP = ? · 60%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 = 3300 kg/hxF = 11 %xP = 60 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 140 °C110 °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 = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 110 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 150 °C110 °Cwater · 25 °C80 °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 = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 150 → 110 °Ct: 25 → 80 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
380 g of CO₂P = 700 kPaT = 85 °CV = ?

What volume V does 380 g of CO₂ occupy at 700 kPa and 85 °C? Treat it as an ideal gas.

m = 380 gM = 44.01 g/molP = 700 kPaT = 85 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of N₂ goes from 350 kPa, 2 L, 40 °C to 350 kPa and 200 °C. Find the new volume V2.

P1 = 350 kPaV1 = 2 LT1 = 40 °CP2 = 350 kPaT2 = 200 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NaCl)m(NaCl)÷ M(X)× 2/ν× M(NaCl)

8 g of Na and 10 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?

M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 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, 30 g of Fe2O3 reacts completely (the other reactants are in excess) and 17.36 g of CO2 is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/molactual: 17.36 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 183.3 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) = 550·(15 − 20)/(20 − 35) = 183.3 kg/h

Answer

ṁ2 = 183.3 kg/h
2.Mass balance

Answer: P = 605 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 = 3300·11/60 = 605 kg/h

Answer

P = 605 kg/h
3.Energy balance

Answer: tout = 60.02 °C

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

Heat released by the hot stream

Q = 3·2.44·(140 − 110) = 219.6 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 219.6/(1.5·4.18) = 60.02 °C

Answer

tout = 60.02 °C
4.Energy balance

Answer: ṁc = 0.5324 kg/s

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

Heat released

Qh = 1·3.6·(150 − 110) = 144 kW

Heat that reaches the water

Qc = (1 − 0.15)·144 = 122.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 122.4/(4.18·55) = 0.5324 kg/s

Answer

ṁc = 0.5324 kg/s
5.Ideal gas law

Answer: V = 36.73 L

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

Moles

n = 380/44.01 = 8.634 mol
T = 85 + 273.15 = 358.1 K

Volume

V = nRT/P = 8.634·8.314·358.1/700 = 36.73 L

Answer

V = 36.73 L
6.Ideal gas law

Answer: V2 = 3.022 L

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

Absolute temperatures

T1 = 313.1 K, T2 = 473.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(350/350)·(473.1/313.1) = 3.022 L

Answer

V2 = 3.022 L
7.Stoichiometry

Answer: m = 16.49 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(Na) = 8/22.99 = 0.348 mol
n(Cl2) = 10/70.9 = 0.141 mol

Limiting reactant

0.348/2 = 0.174, 0.141/1 = 0.141 → Cl2

Mass of product

n(NaCl) = 0.141·2 = 0.2821 mol
m = 0.2821·58.44 = 16.49 g

Answer

m = 16.49 g
8.Stoichiometry

Answer: Y = 69.99 %

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) = 30/159.69 = 0.1879 mol
m = 0.1879·3/1·44.01 = 24.8 g

Percent yield

Y = 17.36/24.8·100 = 69.99%

Answer

Y = 69.99%

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