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 1400 kg/h · 10% sugarStream 2ṁ₂ = ? · 50% 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 = 400 kg/hx1 = 10 %x2 = 50 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3800 kg/h · 18% sugarProduct PP = ? · 45%Vapour W (pure water)0% sugar

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

F = 3800 kg/hxF = 18 %xP = 45 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 180 °C100 °Cwater · 20 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 180 → 100 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 180 °C150 °Cwater · 10 °C90 °C · ṁ = ?5% lost

Hot glycol mix 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 = 3 kg/scp,h = 3.6 kJ/(kg·K)Th: 180 → 150 °Ct: 10 → 90 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
210 g of HeV = 10 LT = 40 °CP = ?

A tank of 10 L contains 210 g of He at 40 °C. Treat it as an ideal gas and find the pressure P.

m = 210 gM = 4.003 g/molV = 10 LT = 40 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
Ar, M = 39.95 g/molP = 100 kPaT = 150 °Cρ = ?

Find the density ρ of Ar at 100 kPa and 150 °C, treating it as an ideal gas.

M = 39.95 g/molP = 100 kPaT = 150 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

6 g of Mg and 10 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(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 16 g of Al reacts completely (the other reactants are in excess) and 64.84 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 64.84 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 240 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) = 400·(10 − 25)/(25 − 50) = 240 kg/h

Answer

ṁ2 = 240 kg/h
2.Mass balance

Answer: P = 1520 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 = 3800·18/45 = 1520 kg/h

Answer

P = 1520 kg/h
3.Energy balance

Answer: tout = 120.5 °C

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

Heat released by the hot stream

Q = 2.5·2.1·(180 − 100) = 420 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 420/(1·4.18) = 120.5 °C

Answer

tout = 120.5 °C
4.Energy balance

Answer: ṁc = 0.9205 kg/s

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

Heat released

Qh = 3·3.6·(180 − 150) = 324 kW

Heat that reaches the water

Qc = (1 − 0.05)·324 = 307.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 307.8/(4.18·80) = 0.9205 kg/s

Answer

ṁc = 0.9205 kg/s
5.Ideal gas law

Answer: P = 13658 kPa

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

Moles

n = m/M = 210/4.003 = 52.46 mol
T = 40 + 273.15 = 313.1 K

Pressure

P = nRT/V = 52.46·8.314·313.1/0.01 = 13658 kPa

Answer

P = 13658 kPa
6.Ideal gas law

Answer: ρ = 1.136 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 = 100·39.95/(8.314·423.1) = 1.136 kg/m³

Answer

ρ = 1.136 kg/m³
7.Stoichiometry

Answer: m = 9.947 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) = 6/24.31 = 0.2468 mol
n(O2) = 10/32 = 0.3125 mol

Limiting reactant

0.2468/2 = 0.1234, 0.3125/1 = 0.3125 → Mg

Mass of product

n(MgO) = 0.1234·2 = 0.2468 mol
m = 0.2468·40.3 = 9.947 g

Answer

m = 9.947 g
8.Stoichiometry

Answer: Y = 82 %

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(Al) = 16/26.98 = 0.593 mol
m = 0.593·2/2·133.33 = 79.07 g

Percent yield

Y = 64.84/79.07·100 = 82%

Answer

Y = 82%

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