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 1650 kg/h · 20% sugarStream 2ṁ₂ = ? · 40% sugarProduct25% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 25% sugar? Find ṁ2.

ṁ1 = 650 kg/hx1 = 20 %x2 = 40 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4200 kg/h · 10% NaOHProduct P40% 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 = 4200 kg/hxF = 10 %xP = 40 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 140 °C70 °Cwater · 10 °Ct = ? °C

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

ṁh = 1 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 70 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 140 °C70 °Cwater · 10 °C85 °C · ṁ = ?5% lost

Hot ethanol 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 = 1.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 70 °Ct: 10 → 85 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
440 g of CO₂V = 5 LT = 0 °CP = ?

A tank of 5 L contains 440 g of CO₂ at 0 °C. Treat it as an ideal gas and find the pressure P.

m = 440 gM = 44.01 g/molV = 5 LT = 0 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 250 kPaT = 150 °Cρ = ?

Find the density ρ of N₂ at 250 kPa and 150 °C, treating it as an ideal gas.

M = 28.01 g/molP = 250 kPaT = 150 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(Fe)m(Fe)÷ M(X)× 2/ν× M(Fe)

8 g of Fe2O3 and 40 g of CO react according to Fe2O3 + 3CO → 2Fe + 3CO2. What mass m of Fe can form?

M(Fe2O3) = 159.69M(CO) = 28.01M(Fe) = 55.85 g/mol
m = g
8.StoichiometryHard
m(N2​)n(N2​)n(NH3​)m(NH3​)÷ M(N2​)× 2/1× M(NH3​)

In the reaction N2 + 3H2 → 2NH3, 16 g of N2 reacts completely (the other reactants are in excess) and 17.7 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 17.7 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 216.7 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
sugar: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 650·(20 − 25)/(25 − 40) = 216.7 kg/h

Answer

ṁ2 = 216.7 kg/h
2.Mass balance

Answer: W = 3150 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 = 4200·10/40 = 1050 kg/h

Total balance

W = F − P = 4200 − 1050 = 3150 kg/h

Answer

W = 3150 kg/h
3.Energy balance

Answer: tout = 30.43 °C

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

Heat released by the hot stream

Q = 1·2.44·(140 − 70) = 170.8 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 170.8/(2·4.18) = 30.43 °C

Answer

tout = 30.43 °C
4.Energy balance

Answer: ṁc = 0.7764 kg/s

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

Heat released

Qh = 1.5·2.44·(140 − 70) = 256.2 kW

Heat that reaches the water

Qc = (1 − 0.05)·256.2 = 243.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 243.4/(4.18·75) = 0.7764 kg/s

Answer

ṁc = 0.7764 kg/s
5.Ideal gas law

Answer: P = 4541 kPa

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

Moles

n = m/M = 440/44.01 = 9.998 mol
T = 0 + 273.15 = 273.1 K

Pressure

P = nRT/V = 9.998·8.314·273.1/0.005 = 4541 kPa

Answer

P = 4541 kPa
6.Ideal gas law

Answer: ρ = 1.99 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 150 + 273.15 = 423.1 K

Density

ρ = 250·28.01/(8.314·423.1) = 1.99 kg/m³

Answer

ρ = 1.99 kg/m³
7.Stoichiometry

Answer: m = 5.596 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(Fe2O3) = 8/159.69 = 0.0501 mol
n(CO) = 40/28.01 = 1.428 mol

Limiting reactant

0.0501/1 = 0.0501, 1.428/3 = 0.476 → Fe2O3

Mass of product

n(Fe) = 0.0501·2 = 0.1002 mol
m = 0.1002·55.85 = 5.596 g

Answer

m = 5.596 g
8.Stoichiometry

Answer: Y = 91.01 %

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(N2) = 16/28.02 = 0.571 mol
m = 0.571·2/1·17.03 = 19.45 g

Percent yield

Y = 17.7/19.45·100 = 91.01%

Answer

Y = 91.01%

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