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 1850 kg/h · 10% NaOHStream 2ṁ₂ = ? · 30% NaOHProduct20% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 20% NaOH? Find ṁ2.

ṁ1 = 850 kg/hx1 = 10 %x2 = 30 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3100 kg/h · 10% NaOHProduct P50% 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 = 3100 kg/hxF = 10 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 120 °C70 °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 = 1 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 70 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 170 °C60 °Cwater · 30 °C70 °C · ṁ = ?10% lost

Hot ethanol heats water in an exchanger that loses 10% 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: 170 → 60 °Ct: 30 → 70 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
300 g of O₂V = 20 LT = 30 °CP = ?

A tank of 20 L contains 300 g of O₂ at 30 °C. Treat it as an ideal gas and find the pressure P.

m = 300 gM = 32 g/molV = 20 LT = 30 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of N₂ goes from 300 kPa, 5 L, 35 °C to 250 kPa and 100 °C. Find the new volume V2.

P1 = 300 kPaV1 = 5 LT1 = 35 °CP2 = 250 kPaT2 = 100 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

30 g of Mg and 20 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(Na)n(Na)n(NaCl)m(NaCl)÷ M(Na)× 2/2× M(NaCl)

In the reaction 2Na + Cl2 → 2NaCl, 12 g of Na reacts completely (the other reactants are in excess) and 22.88 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 22.88 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 850 kg/h

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

Balances

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

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 850·(10 − 20)/(20 − 30) = 850 kg/h

Answer

ṁ2 = 850 kg/h
2.Mass balance

Answer: W = 2480 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 = 3100·10/50 = 620 kg/h

Total balance

W = F − P = 3100 − 620 = 2480 kg/h

Answer

W = 2480 kg/h
3.Energy balance

Answer: tout = 39.59 °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·(120 − 70) = 122 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 122/(2·4.18) = 39.59 °C

Answer

tout = 39.59 °C
4.Energy balance

Answer: ṁc = 4.334 kg/s

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

Heat released

Qh = 3·2.44·(170 − 60) = 805.2 kW

Heat that reaches the water

Qc = (1 − 0.1)·805.2 = 724.7 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 724.7/(4.18·40) = 4.334 kg/s

Answer

ṁc = 4.334 kg/s
5.Ideal gas law

Answer: P = 1181 kPa

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

Moles

n = m/M = 300/32 = 9.375 mol
T = 30 + 273.15 = 303.1 K

Pressure

P = nRT/V = 9.375·8.314·303.1/0.02 = 1181 kPa

Answer

P = 1181 kPa
6.Ideal gas law

Answer: V2 = 7.266 L

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

Absolute temperatures

T1 = 308.1 K, T2 = 373.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(300/250)·(373.1/308.1) = 7.266 L

Answer

V2 = 7.266 L
7.Stoichiometry

Answer: m = 49.73 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) = 30/24.31 = 1.234 mol
n(O2) = 20/32 = 0.625 mol

Limiting reactant

1.234/2 = 0.617, 0.625/1 = 0.625 → Mg

Mass of product

n(MgO) = 0.617·2 = 1.234 mol
m = 1.234·40.3 = 49.73 g

Answer

m = 49.73 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(Na) = 12/22.99 = 0.522 mol
m = 0.522·2/2·58.44 = 30.5 g

Percent yield

Y = 22.88/30.5·100 = 75.01%

Answer

Y = 75.01%

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