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 1500 kg/h · 10% saltStream 2ṁ₂ = ? · 45% saltProduct15% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 15% salt? Find ṁ2.

ṁ1 = 500 kg/hx1 = 10 %x2 = 45 %x3 = 15 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2300 kg/h · 12% sugarProduct P30% sugarVapour W (pure water)W = ? kg/h

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

F = 2300 kg/hxF = 12 %xP = 30 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 170 °C140 °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 = 0.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 140 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 130 °C60 °Cwater · 20 °C55 °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 = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 130 → 60 °Ct: 20 → 55 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
320 g of O₂P = 150 kPaT = 75 °CV = ?

What volume V does 320 g of O₂ occupy at 150 kPa and 75 °C? Treat it as an ideal gas.

m = 320 gM = 32 g/molP = 150 kPaT = 75 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
He, M = 4.003 g/molP = 800 kPaT = 140 °Cρ = ?

Find the density ρ of He at 800 kPa and 140 °C, treating it as an ideal gas.

M = 4.003 g/molP = 800 kPaT = 140 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

12 g of N2 and 16 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?

M(N2) = 28.02M(H2) = 2.016M(NH3) = 17.03 g/mol
m = g
8.StoichiometryHard
m(Fe2​O3​)n(Fe2​O3​)n(Fe)m(Fe)÷ M(Fe2​O3​)× 2/1× M(Fe)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 20 g of Fe2O3 reacts completely (the other reactants are in excess) and 10.49 g of Fe is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(Fe) = 55.85 g/molactual: 10.49 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 83.33 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) = 500·(10 − 15)/(15 − 45) = 83.33 kg/h

Answer

ṁ2 = 83.33 kg/h
2.Mass balance

Answer: W = 1380 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 = 2300·12/30 = 920 kg/h

Total balance

W = F − P = 2300 − 920 = 1380 kg/h

Answer

W = 1380 kg/h
3.Energy balance

Answer: tout = 42.51 °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·(170 − 140) = 36.6 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 36.6/(0.5·4.18) = 42.51 °C

Answer

tout = 42.51 °C
4.Energy balance

Answer: ṁc = 2.196 kg/s

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

Heat released

Qh = 1.5·3.6·(130 − 60) = 378 kW

Heat that reaches the water

Qc = (1 − 0.15)·378 = 321.3 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 321.3/(4.18·35) = 2.196 kg/s

Answer

ṁc = 2.196 kg/s
5.Ideal gas law

Answer: V = 193 L

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

Moles

n = 320/32 = 10 mol
T = 75 + 273.15 = 348.1 K

Volume

V = nRT/P = 10·8.314·348.1/150 = 193 L

Answer

V = 193 L
6.Ideal gas law

Answer: ρ = 0.9323 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 140 + 273.15 = 413.1 K

Density

ρ = 800·4.003/(8.314·413.1) = 0.9323 kg/m³

Answer

ρ = 0.9323 kg/m³
7.Stoichiometry

Answer: m = 14.59 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(N2) = 12/28.02 = 0.4283 mol
n(H2) = 16/2.016 = 7.937 mol

Limiting reactant

0.4283/1 = 0.4283, 7.937/3 = 2.646 → N2

Mass of product

n(NH3) = 0.4283·2 = 0.8565 mol
m = 0.8565·17.03 = 14.59 g

Answer

m = 14.59 g
8.Stoichiometry

Answer: Y = 74.98 %

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) = 20/159.69 = 0.1252 mol
m = 0.1252·2/1·55.85 = 13.99 g

Percent yield

Y = 10.49/13.99·100 = 74.98%

Answer

Y = 74.98%

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