Mass balance
- Steady state
- in = out
- Total balance
- ṁ1 + ṁ2 = ṁ3
- Component balance
- ṁ1x1 + ṁ2x2 = ṁ3x3
- Evaporator
- F·xF = P·xP, W = F − P
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.
1Formulas and methods
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
- Write and balance the chemical equation.
- Convert the given mass to moles: n = m/M.
- Use the coefficients as the mole ratio to find the moles of the substance you want.
- Convert back to mass with m = n·M. With two reactants, find the limiting reactant first.
2Practice problems
1.Mass balanceMedium
How much of stream 2 must be added each hour to stream 1 so that the blended product contains 45% salt? Find ṁ2.
ṁ1 = 950 kg/hx1 = 20 %x2 = 50 %x3 = 45 %
ṁ2 = kg/h
2.Mass balanceHard
An evaporator concentrates a NaOH solution by boiling off pure water. Find the water evaporated W in kg/h.
F = 2100 kg/hxF = 12 %xP = 45 %
W = kg/h
3.Energy balanceMedium
Hot light oil cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature tout.
ṁh = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
Hot light oil 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 = 3 kg/scp,h = 2.1 kJ/(kg·K)Th: 130 → 70 °Ct: 10 → 30 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
What volume V does 300 g of CH₄ occupy at 550 kPa and 0 °C? Treat it as an ideal gas.
m = 300 gM = 16.04 g/molP = 550 kPaT = 0 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
Find the density ρ of O₂ at 950 kPa and 190 °C, treating it as an ideal gas.
M = 32 g/molP = 950 kPaT = 190 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
8 g of Al and 8 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?
M(Al) = 26.98M(Cl2) = 70.9M(AlCl3) = 133.33 g/mol
m = g
8.StoichiometryHard
In the reaction C6H12O6 → 2C2H5OH + 2CO2, 8 g of C6H12O6 reacts completely (the other reactants are in excess) and 3.19 g of C2H5OH is collected. Find the percent yield Y.
M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/molactual: 3.19 g
Y = %
3Solutions
1.Mass balance
Answer: ṁ2 = 4750 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) = 950·(20 − 45)/(45 − 50) = 4750 kg/h
Answer
ṁ2 = 4750 kg/h
2.Mass balance
Answer: W = 1540 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 = 2100·12/45 = 560 kg/h
Total balance
W = F − P = 2100 − 560 = 1540 kg/h
Answer
W = 1540 kg/h
3.Energy balance
Answer: tout = 96.99 °C
Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).
Heat released by the hot stream
Q = 2.5·2.1·(140 − 60) = 420 kW
Heat picked up by the water
tout = tin + Q/(ṁccp,c) = 30 + 420/(1.5·4.18) = 96.99 °C
Answer
tout = 96.99 °C
4.Energy balance
Answer: ṁc = 4.295 kg/s
Energy balance with losses: heat released = heat absorbed + heat lost.
Heat released
Qh = 3·2.1·(130 − 70) = 378 kW
Heat that reaches the water
Qc = (1 − 0.05)·378 = 359.1 kW
Water flow rate
ṁc = Qc/(cp,c·Δt) = 359.1/(4.18·20) = 4.295 kg/s
Answer
ṁc = 4.295 kg/s
5.Ideal gas law
Answer: V = 77.23 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 300/16.04 = 18.7 mol
T = 0 + 273.15 = 273.1 K
Volume
V = nRT/P = 18.7·8.314·273.1/550 = 77.23 L
Answer
V = 77.23 L
6.Ideal gas law
Answer: ρ = 7.895 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 190 + 273.15 = 463.1 K
Density
ρ = 950·32/(8.314·463.1) = 7.895 kg/m³
Answer
ρ = 7.895 kg/m³
7.Stoichiometry
Answer: m = 10.03 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(Al) = 8/26.98 = 0.2965 mol
n(Cl2) = 8/70.9 = 0.1128 mol
Limiting reactant
0.2965/2 = 0.1483, 0.1128/3 = 0.03761 → Cl2
Mass of product
n(AlCl3) = 0.03761·2 = 0.07522 mol
m = 0.07522·133.33 = 10.03 g
Answer
m = 10.03 g
8.Stoichiometry
Answer: Y = 77.97 %
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 = 0.0444·2/1·46.07 = 4.091 g
Percent yield
Y = 3.19/4.091·100 = 77.97%
Answer
Y = 77.97%
ExamProblems · examproblems.com/courses/material-and-energy-balances