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 1400 kg/h · 5% saltStream 2ṁ₂ = ? · 40% 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 = 400 kg/hx1 = 5 %x2 = 40 %x3 = 10 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1300 kg/h · 6% NaOHProduct P40% 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 = 1300 kg/hxF = 6 %xP = 40 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 150 °C90 °Cwater · 15 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 2.1 kJ/(kg·K)Th: 150 → 90 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 170 °C140 °Cwater · 25 °C90 °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 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 140 °Ct: 25 → 90 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
120 g of CO₂V = 30 LT = 65 °CP = ?

A tank of 30 L contains 120 g of CO₂ at 65 °C. Treat it as an ideal gas and find the pressure P.

m = 120 gM = 44.01 g/molV = 30 LT = 65 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 500 kPaT = 100 °Cρ = ?

Find the density ρ of N₂ at 500 kPa and 100 °C, treating it as an ideal gas.

M = 28.01 g/molP = 500 kPaT = 100 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

12 g of Mg and 10 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(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, 8 g of C6H12O6 reacts completely (the other reactants are in excess) and 3.05 g of CO2 is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(CO2) = 44.01 g/molactual: 3.05 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 66.67 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) = 400·(5 − 10)/(10 − 40) = 66.67 kg/h

Answer

ṁ2 = 66.67 kg/h
2.Mass balance

Answer: W = 1105 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 = 1300·6/40 = 195 kg/h

Total balance

W = F − P = 1300 − 195 = 1105 kg/h

Answer

W = 1105 kg/h
3.Energy balance

Answer: tout = 55.19 °C

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

Heat released by the hot stream

Q = 2·2.1·(150 − 90) = 252 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 252/(1.5·4.18) = 55.19 °C

Answer

tout = 55.19 °C
4.Energy balance

Answer: ṁc = 0.2425 kg/s

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

Heat released

Qh = 1·2.44·(170 − 140) = 73.2 kW

Heat that reaches the water

Qc = (1 − 0.1)·73.2 = 65.88 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 65.88/(4.18·65) = 0.2425 kg/s

Answer

ṁc = 0.2425 kg/s
5.Ideal gas law

Answer: P = 255.5 kPa

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

Moles

n = m/M = 120/44.01 = 2.727 mol
T = 65 + 273.15 = 338.1 K

Pressure

P = nRT/V = 2.727·8.314·338.1/0.03 = 255.5 kPa

Answer

P = 255.5 kPa
6.Ideal gas law

Answer: ρ = 4.514 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 100 + 273.15 = 373.1 K

Density

ρ = 500·28.01/(8.314·373.1) = 4.514 kg/m³

Answer

ρ = 4.514 kg/m³
7.Stoichiometry

Answer: m = 19.89 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) = 12/24.31 = 0.4936 mol
n(O2) = 10/32 = 0.3125 mol

Limiting reactant

0.4936/2 = 0.2468, 0.3125/1 = 0.3125 → Mg

Mass of product

n(MgO) = 0.2468·2 = 0.4936 mol
m = 0.4936·40.3 = 19.89 g

Answer

m = 19.89 g
8.Stoichiometry

Answer: Y = 78.03 %

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) = 8/180.16 = 0.0444 mol
m = 0.0444·2/1·44.01 = 3.909 g

Percent yield

Y = 3.05/3.909·100 = 78.03%

Answer

Y = 78.03%

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