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 · 35% NaOHStream 2ṁ₂ = ? · 15% 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 = 500 kg/hx1 = 35 %x2 = 15 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1700 kg/h · 9% 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 = 1700 kg/hxF = 9 %xP = 45 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 120 °C80 °Cwater · 10 °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: 120 → 80 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 140 °C70 °Cwater · 25 °C80 °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 = 1.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 70 °Ct: 25 → 80 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
30 g of N₂V = 50 LT = 85 °CP = ?

A tank of 50 L contains 30 g of N₂ at 85 °C. Treat it as an ideal gas and find the pressure P.

m = 30 gM = 28.01 g/molV = 50 LT = 85 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of Ar goes from 350 kPa, 2 L, 10 °C to 600 kPa and 100 °C. Find the new volume V2.

P1 = 350 kPaV1 = 2 LT1 = 10 °CP2 = 600 kPaT2 = 100 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

8 g of N2 and 40 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(C6​H12​O6​)n(C6​H12​O6​)n(CO2​)m(CO2​)÷ M(C6​H12​O6​)× 2/1× M(CO2​)

In the reaction C6H12O6 → 2C2H5OH + 2CO2, 10 g of C6H12O6 reacts completely (the other reactants are in excess) and 3.66 g of CO2 is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(CO2) = 44.01 g/molactual: 3.66 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1500 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) = 500·(35 − 20)/(20 − 15) = 1500 kg/h

Answer

ṁ2 = 1500 kg/h
2.Mass balance

Answer: P = 340 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 = 1700·9/45 = 340 kg/h

Answer

P = 340 kg/h
3.Energy balance

Answer: tout = 33.35 °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·(120 − 80) = 48.8 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 48.8/(0.5·4.18) = 33.35 °C

Answer

tout = 33.35 °C
4.Energy balance

Answer: ṁc = 1.003 kg/s

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

Heat released

Qh = 1.5·2.44·(140 − 70) = 256.2 kW

Heat that reaches the water

Qc = (1 − 0.1)·256.2 = 230.6 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 230.6/(4.18·55) = 1.003 kg/s

Answer

ṁc = 1.003 kg/s
5.Ideal gas law

Answer: P = 63.78 kPa

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

Moles

n = m/M = 30/28.01 = 1.071 mol
T = 85 + 273.15 = 358.1 K

Pressure

P = nRT/V = 1.071·8.314·358.1/0.05 = 63.78 kPa

Answer

P = 63.78 kPa
6.Ideal gas law

Answer: V2 = 1.537 L

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

Absolute temperatures

T1 = 283.1 K, T2 = 373.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(350/600)·(373.1/283.1) = 1.537 L

Answer

V2 = 1.537 L
7.Stoichiometry

Answer: m = 9.724 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) = 8/28.02 = 0.2855 mol
n(H2) = 40/2.016 = 19.84 mol

Limiting reactant

0.2855/1 = 0.2855, 19.84/3 = 6.614 → N2

Mass of product

n(NH3) = 0.2855·2 = 0.571 mol
m = 0.571·17.03 = 9.724 g

Answer

m = 9.724 g
8.Stoichiometry

Answer: Y = 74.91 %

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(C6H12O6) = 10/180.16 = 0.05551 mol
m = 0.05551·2/1·44.01 = 4.886 g

Percent yield

Y = 3.66/4.886·100 = 74.91%

Answer

Y = 74.91%

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