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 1600 kg/h · 20% saltStream 2ṁ₂ = ? · 35% saltProduct30% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% salt? Find ṁ2.

ṁ1 = 600 kg/hx1 = 20 %x2 = 35 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1000 kg/h · 9% NaOHProduct PP = ? · 55%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 = 1000 kg/hxF = 9 %xP = 55 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 140 °C90 °Cwater · 30 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 90 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 150 °C100 °Cwater · 10 °C25 °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 = 0.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 150 → 100 °Ct: 10 → 25 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
200 g of N₂V = 30 LT = 80 °CP = ?

A tank of 30 L contains 200 g of N₂ at 80 °C. Treat it as an ideal gas and find the pressure P.

m = 200 gM = 28.01 g/molV = 30 LT = 80 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of O₂ goes from 400 kPa, 10 L, 35 °C to 250 kPa and 240 °C. Find the new volume V2.

P1 = 400 kPaV1 = 10 LT1 = 35 °CP2 = 250 kPaT2 = 240 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

4 g of Mg and 64 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(CH4​)n(CH4​)n(CO2​)m(CO2​)÷ M(CH4​)× 1/1× M(CO2​)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 6 g of CH4 reacts completely (the other reactants are in excess) and 10.21 g of CO2 is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(CO2) = 44.01 g/molactual: 10.21 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1200 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) = 600·(20 − 30)/(30 − 35) = 1200 kg/h

Answer

ṁ2 = 1200 kg/h
2.Mass balance

Answer: P = 163.6 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 = 1000·9/55 = 163.6 kg/h

Answer

P = 163.6 kg/h
3.Energy balance

Answer: tout = 101.8 °C

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

Heat released by the hot stream

Q = 2.5·3.6·(140 − 90) = 450 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 450/(1.5·4.18) = 101.8 °C

Answer

tout = 101.8 °C
4.Energy balance

Answer: ṁc = 1.364 kg/s

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

Heat released

Qh = 0.5·3.6·(150 − 100) = 90 kW

Heat that reaches the water

Qc = (1 − 0.05)·90 = 85.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 85.5/(4.18·15) = 1.364 kg/s

Answer

ṁc = 1.364 kg/s
5.Ideal gas law

Answer: P = 698.8 kPa

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

Moles

n = m/M = 200/28.01 = 7.14 mol
T = 80 + 273.15 = 353.1 K

Pressure

P = nRT/V = 7.14·8.314·353.1/0.03 = 698.8 kPa

Answer

P = 698.8 kPa
6.Ideal gas law

Answer: V2 = 26.64 L

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

Absolute temperatures

T1 = 308.1 K, T2 = 513.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(400/250)·(513.1/308.1) = 26.64 L

Answer

V2 = 26.64 L
7.Stoichiometry

Answer: m = 6.631 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) = 4/24.31 = 0.1645 mol
n(O2) = 64/32 = 2 mol

Limiting reactant

0.1645/2 = 0.08227, 2/1 = 2 → Mg

Mass of product

n(MgO) = 0.08227·2 = 0.1645 mol
m = 0.1645·40.3 = 6.631 g

Answer

m = 6.631 g
8.Stoichiometry

Answer: Y = 62.02 %

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(CH4) = 6/16.04 = 0.3741 mol
m = 0.3741·1/1·44.01 = 16.46 g

Percent yield

Y = 10.21/16.46·100 = 62.02%

Answer

Y = 62.02%

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