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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 1900 kg/h · 5% saltStream 2ṁ₂ = ? · 30% saltProduct10% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 10% salt? Find ṁ2.

ṁ1 = 900 kg/hx1 = 5 %x2 = 30 %x3 = 10 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4100 kg/h · 16% NaOHProduct PP = ? · 60%Vapour W (pure water)0% NaOH

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

F = 4100 kg/hxF = 16 %xP = 60 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 120 °C60 °Cwater · 10 °Ct = ? °C

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

ṁh = 3 kg/scp,h = 3.6 kJ/(kg·K)Th: 120 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 120 °C80 °Cwater · 10 °C75 °C · ṁ = ?15% lost

Hot ethanol 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 = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 80 °Ct: 10 → 75 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
170 g of ArP = 800 kPaT = 0 °CV = ?

What volume V does 170 g of Ar occupy at 800 kPa and 0 °C? Treat it as an ideal gas.

m = 170 gM = 39.95 g/molP = 800 kPaT = 0 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 950 kPaT = 70 °Cρ = ?

Find the density ρ of N₂ at 950 kPa and 70 °C, treating it as an ideal gas.

M = 28.01 g/molP = 950 kPaT = 70 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 4/ν× M(H2​O)

30 g of C3H8 and 10 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of H2O can form?

M(C3H8) = 44.1M(O2) = 32M(H2O) = 18.02 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, 30 g of Fe2O3 reacts completely (the other reactants are in excess) and 15.74 g of Fe is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(Fe) = 55.85 g/molactual: 15.74 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 225 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) = 900·(5 − 10)/(10 − 30) = 225 kg/h

Answer

ṁ2 = 225 kg/h
2.Mass balance

Answer: P = 1093 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 = 4100·16/60 = 1093 kg/h

Answer

P = 1093 kg/h
3.Energy balance

Answer: tout = 113.3 °C

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

Heat released by the hot stream

Q = 3·3.6·(120 − 60) = 648 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 648/(1.5·4.18) = 113.3 °C

Answer

tout = 113.3 °C
4.Energy balance

Answer: ṁc = 0.916 kg/s

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

Heat released

Qh = 3·2.44·(120 − 80) = 292.8 kW

Heat that reaches the water

Qc = (1 − 0.15)·292.8 = 248.9 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 248.9/(4.18·65) = 0.916 kg/s

Answer

ṁc = 0.916 kg/s
5.Ideal gas law

Answer: V = 12.08 L

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

Moles

n = 170/39.95 = 4.255 mol
T = 0 + 273.15 = 273.1 K

Volume

V = nRT/P = 4.255·8.314·273.1/800 = 12.08 L

Answer

V = 12.08 L
6.Ideal gas law

Answer: ρ = 9.327 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 70 + 273.15 = 343.1 K

Density

ρ = 950·28.01/(8.314·343.1) = 9.327 kg/m³

Answer

ρ = 9.327 kg/m³
7.Stoichiometry

Answer: m = 4.505 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(C3H8) = 30/44.1 = 0.6803 mol
n(O2) = 10/32 = 0.3125 mol

Limiting reactant

0.6803/1 = 0.6803, 0.3125/5 = 0.0625 → O2

Mass of product

n(H2O) = 0.0625·4 = 0.25 mol
m = 0.25·18.02 = 4.505 g

Answer

m = 4.505 g
8.Stoichiometry

Answer: Y = 75.01 %

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(Fe2O3) = 30/159.69 = 0.1879 mol
m = 0.1879·2/1·55.85 = 20.98 g

Percent yield

Y = 15.74/20.98·100 = 75.01%

Answer

Y = 75.01%

ExamProblems · examproblems.com/courses/material-and-energy-balances