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10 FE practice problems: combustion CO₂ and air, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten original problems practice combustion CO₂ and air, a topic from the FE Environmental exam specification, using the Thermodynamics, Combustion Processes part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.

How to use these problems

Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.

The problems

Problem 1 · FE Environmental, Combustion CO₂ and air

When 315 kg of methane (CH4) burns completely in air, using atomic weights C = 12.01, H = 1.008 and O = 16.00, the mass of CO2 produced is most nearly:

  • A 1,730 kg
  • B 864 kg
  • C 236 kg
  • D 1,150 kg

Handbook: Thermodynamics, Combustion Processes, FE Reference Handbook 10.6

Show the worked solution
Answer
B (864 kg)
Given
m = 315 kg, n = 1, awC = 12.01, awH = 1.008, awO = 16.00, h = 4
Find
mass of CO2 (kg)
Handbook
Thermodynamics, Combustion Processes, page 153
Equation
CnH2n+2 + (3n+1)/2 O2 → n CO2 + (n+1) H2O; mass ratio CO2/fuel = n·MW(CO2)/MW(fuel)
Substitute
  1. CH4 + 2 O2 → 1 CO2 + 2 H2O
  2. MW fuel = 1(12.01) + 4(1.008) = 16.04; MW CO2 = 12.01 + 2(16.00) = 44.01
  3. m CO2 = 315 × 1 × 44.01/16.04 = 864 kg
Result
864 kg, 3 significant figures
Check
all the carbon ends up in CO2: the carbon in the fuel equals 12.01/44.01 of the CO2 mass.
Why the others are wrong
  • A: used the moles of water (n + 1) for CO2
  • C: is the mass of carbon in the fuel, not of CO2
  • D: treated the fuel mass as carbon only (hydrogen ignored)

Problem 2 · FE Environmental, Combustion CO₂ and air

When 910 lb of ethane (C2H6) burns completely in air, using atomic weights C = 12.01, H = 1.008 and O = 16.00, the mass of CO2 produced is most nearly:

  • A 1,330 lb
  • B 3,330 lb
  • C 2,660 lb
  • D 1,940 lb

Handbook: Thermodynamics, Combustion Processes, FE Reference Handbook 10.6

Show the worked solution
Answer
C (2,660 lb)
Given
m = 910 lb, n = 2, awC = 12.01, awH = 1.008, awO = 16.00, h = 6
Find
mass of CO2 (lb)
Handbook
Thermodynamics, Combustion Processes, page 153
Equation
CnH2n+2 + (3n+1)/2 O2 → n CO2 + (n+1) H2O; mass ratio CO2/fuel = n·MW(CO2)/MW(fuel)
Substitute
  1. C2H6 + 3.5 O2 → 2 CO2 + 3 H2O
  2. MW fuel = 2(12.01) + 6(1.008) = 30.07; MW CO2 = 12.01 + 2(16.00) = 44.01
  3. m CO2 = 910 × 2 × 44.01/30.07 = 2,660 lb
Result
2,660 lb, 3 significant figures
Check
all the carbon ends up in CO2: the carbon in the fuel equals 12.01/44.01 of the CO2 mass.
Why the others are wrong
  • A: used 1 mol of CO2 per mol of fuel instead of n
  • B: treated the fuel mass as carbon only (hydrogen ignored)
  • D: counted only the oxygen in the CO2

Problem 3 · FE Environmental, Combustion CO₂ and air

When 420 lb of ethane (C2H6) burns completely in air, using atomic weights C = 12.01, H = 1.008 and O = 16.00, the mass of CO2 produced is most nearly:

  • A 336 lb
  • B 615 lb
  • C 1,230 lb
  • D 894 lb

Handbook: Thermodynamics, Combustion Processes, FE Reference Handbook 10.6

Show the worked solution
Answer
C (1,230 lb)
Given
m = 420 lb, n = 2, awC = 12.01, awH = 1.008, awO = 16.00, h = 6
Find
mass of CO2 (lb)
Handbook
Thermodynamics, Combustion Processes, page 153
Equation
CnH2n+2 + (3n+1)/2 O2 → n CO2 + (n+1) H2O; mass ratio CO2/fuel = n·MW(CO2)/MW(fuel)
Substitute
  1. C2H6 + 3.5 O2 → 2 CO2 + 3 H2O
  2. MW fuel = 2(12.01) + 6(1.008) = 30.07; MW CO2 = 12.01 + 2(16.00) = 44.01
  3. m CO2 = 420 × 2 × 44.01/30.07 = 1,230 lb
Result
1,230 lb, 3 significant figures
Check
all the carbon ends up in CO2: the carbon in the fuel equals 12.01/44.01 of the CO2 mass.
Why the others are wrong
  • A: is the mass of carbon in the fuel, not of CO2
  • B: used 1 mol of CO2 per mol of fuel instead of n
  • D: counted only the oxygen in the CO2

Problem 4 · FE Environmental, Combustion CO₂ and air

Methane (CH4) burns completely with 10% excess air. Taking air as 3.76 mol N2 per mol O2, the moles of air supplied per mole of fuel are most nearly:

  • A 10.5 mol air/mol fuel
  • B 2.20 mol air/mol fuel
  • C 15.7 mol air/mol fuel
  • D 9.52 mol air/mol fuel

Handbook: Thermodynamics, Combustion Processes, FE Reference Handbook 10.6

Show the worked solution
Answer
A (10.5 mol air/mol fuel)
Given
n = 1, excess = 10%, n2 = 3.76, h = 4, two = 2
Find
moles of air per mole of fuel
Handbook
Thermodynamics, Combustion Processes (combustion in air, excess air), page 153
Equation
O2 needed = (3n + 1)/2 per mol CnH2n+2; air = 4.76 × O2 × (1 + excess)
Substitute
  1. CH4 + 2(O2 + 3.76 N2) → 1 CO2 + 2 H2O + 7.520 N2 (theoretical)
  2. Theoretical air = 2 × 4.76 = 9.520 mol/mol fuel
  3. Supplied air = 9.520 × (1 + 10/100) = 10.5 mol air/mol fuel
Result
10.5 mol air/mol fuel, 3 significant figures
Check
excess oxygen leaves in the flue gas; the nitrogen passes through unchanged.
Why the others are wrong
  • B: is the oxygen, not the air (nitrogen left out)
  • C: counted the moles of products instead of the O2 needed
  • D: left out the excess air

Problem 5 · FE Environmental, Combustion CO₂ and air

When 550 lb of butane (C4H10) burns completely in air, using atomic weights C = 12.01, H = 1.008 and O = 16.00, the mass of CO2 produced is most nearly:

  • A 2,020 lb
  • B 1,210 lb
  • C 1,670 lb
  • D 455 lb

Handbook: Thermodynamics, Combustion Processes, FE Reference Handbook 10.6

Show the worked solution
Answer
C (1,670 lb)
Given
m = 550 lb, n = 4, awC = 12.01, awH = 1.008, awO = 16.00, h = 10
Find
mass of CO2 (lb)
Handbook
Thermodynamics, Combustion Processes, page 153
Equation
CnH2n+2 + (3n+1)/2 O2 → n CO2 + (n+1) H2O; mass ratio CO2/fuel = n·MW(CO2)/MW(fuel)
Substitute
  1. C4H10 + 6.5 O2 → 4 CO2 + 5 H2O
  2. MW fuel = 4(12.01) + 10(1.008) = 58.12; MW CO2 = 12.01 + 2(16.00) = 44.01
  3. m CO2 = 550 × 4 × 44.01/58.12 = 1,670 lb
Result
1,670 lb, 3 significant figures
Check
all the carbon ends up in CO2: the carbon in the fuel equals 12.01/44.01 of the CO2 mass.
Why the others are wrong
  • A: treated the fuel mass as carbon only (hydrogen ignored)
  • B: counted only the oxygen in the CO2
  • D: is the mass of carbon in the fuel, not of CO2

Problem 6 · FE Environmental, Combustion CO₂ and air

Propane (C3H8) burns completely with 50% excess air. Taking air as 3.76 mol N2 per mol O2, the moles of air supplied per mole of fuel are most nearly:

  • A 23.8 mol air/mol fuel
  • B 28.2 mol air/mol fuel
  • C 50.0 mol air/mol fuel
  • D 35.7 mol air/mol fuel

Handbook: Thermodynamics, Combustion Processes, FE Reference Handbook 10.6

Show the worked solution
Answer
D (35.7 mol air/mol fuel)
Given
n = 3, excess = 50%, n2 = 3.76, h = 8, two = 2
Find
moles of air per mole of fuel
Handbook
Thermodynamics, Combustion Processes (combustion in air, excess air), page 153
Equation
O2 needed = (3n + 1)/2 per mol CnH2n+2; air = 4.76 × O2 × (1 + excess)
Substitute
  1. C3H8 + 5(O2 + 3.76 N2) → 3 CO2 + 4 H2O + 18.80 N2 (theoretical)
  2. Theoretical air = 5 × 4.76 = 23.80 mol/mol fuel
  3. Supplied air = 23.80 × (1 + 50/100) = 35.7 mol air/mol fuel
Result
35.7 mol air/mol fuel, 3 significant figures
Check
excess oxygen leaves in the flue gas; the nitrogen passes through unchanged.
Why the others are wrong
  • A: left out the excess air
  • B: used 3.76 (the nitrogen) instead of 4.76 moles of air per mole of O2
  • C: counted the moles of products instead of the O2 needed

Problem 7 · FE Environmental, Combustion CO₂ and air

Methane (CH4) burns completely with 15% excess air. Taking air as 3.76 mol N2 per mol O2, the moles of air supplied per mole of fuel are most nearly:

  • A 8.65 mol air/mol fuel
  • B 9.52 mol air/mol fuel
  • C 8.21 mol air/mol fuel
  • D 10.9 mol air/mol fuel

Handbook: Thermodynamics, Combustion Processes, FE Reference Handbook 10.6

Show the worked solution
Answer
D (10.9 mol air/mol fuel)
Given
n = 1, excess = 15%, n2 = 3.76, h = 4, two = 2
Find
moles of air per mole of fuel
Handbook
Thermodynamics, Combustion Processes (combustion in air, excess air), page 153
Equation
O2 needed = (3n + 1)/2 per mol CnH2n+2; air = 4.76 × O2 × (1 + excess)
Substitute
  1. CH4 + 2(O2 + 3.76 N2) → 1 CO2 + 2 H2O + 7.520 N2 (theoretical)
  2. Theoretical air = 2 × 4.76 = 9.520 mol/mol fuel
  3. Supplied air = 9.520 × (1 + 15/100) = 10.9 mol air/mol fuel
Result
10.9 mol air/mol fuel, 3 significant figures
Check
excess oxygen leaves in the flue gas; the nitrogen passes through unchanged.
Why the others are wrong
  • A: used 3.76 (the nitrogen) instead of 4.76 moles of air per mole of O2
  • B: left out the excess air
  • C: counted half the hydrogen's oxygen demand

Problem 8 · FE Environmental, Combustion CO₂ and air

Methane (CH4) burns completely with 20% excess air. Taking air as 3.76 mol N2 per mol O2, the moles of air supplied per mole of fuel are most nearly:

  • A 8.57 mol air/mol fuel
  • B 11.4 mol air/mol fuel
  • C 9.52 mol air/mol fuel
  • D 9.02 mol air/mol fuel

Handbook: Thermodynamics, Combustion Processes, FE Reference Handbook 10.6

Show the worked solution
Answer
B (11.4 mol air/mol fuel)
Given
n = 1, excess = 20%, n2 = 3.76, h = 4, two = 2
Find
moles of air per mole of fuel
Handbook
Thermodynamics, Combustion Processes (combustion in air, excess air), page 153
Equation
O2 needed = (3n + 1)/2 per mol CnH2n+2; air = 4.76 × O2 × (1 + excess)
Substitute
  1. CH4 + 2(O2 + 3.76 N2) → 1 CO2 + 2 H2O + 7.520 N2 (theoretical)
  2. Theoretical air = 2 × 4.76 = 9.520 mol/mol fuel
  3. Supplied air = 9.520 × (1 + 20/100) = 11.4 mol air/mol fuel
Result
11.4 mol air/mol fuel, 3 significant figures
Check
excess oxygen leaves in the flue gas; the nitrogen passes through unchanged.
Why the others are wrong
  • A: counted half the hydrogen's oxygen demand
  • C: left out the excess air
  • D: used 3.76 (the nitrogen) instead of 4.76 moles of air per mole of O2

Problem 9 · FE Environmental, Combustion CO₂ and air

Ethane (C2H6) burns completely with stoichiometric (theoretical) air. Taking air as 3.76 mol N2 per mol O2, the moles of air supplied per mole of fuel are most nearly:

  • A 16.7 mol air/mol fuel
  • B 3.50 mol air/mol fuel
  • C 11.9 mol air/mol fuel
  • D 23.8 mol air/mol fuel

Handbook: Thermodynamics, Combustion Processes, FE Reference Handbook 10.6

Show the worked solution
Answer
A (16.7 mol air/mol fuel)
Given
n = 2, excess = 0%, n2 = 3.76, h = 6, two = 2
Find
moles of air per mole of fuel
Handbook
Thermodynamics, Combustion Processes (combustion in air, excess air), page 153
Equation
O2 needed = (3n + 1)/2 per mol CnH2n+2; air = 4.76 × O2 × (1 + excess)
Substitute
  1. C2H6 + 3.5(O2 + 3.76 N2) → 2 CO2 + 3 H2O + 13.16 N2 (theoretical)
  2. Theoretical air = 3.5 × 4.76 = 16.66 mol/mol fuel
  3. Supplied air = 16.66 × (1 + 0/100) = 16.7 mol air/mol fuel
Result
16.7 mol air/mol fuel, 3 significant figures
Check
excess oxygen leaves in the flue gas; the nitrogen passes through unchanged.
Why the others are wrong
  • B: is the oxygen, not the air (nitrogen left out)
  • C: counted half the hydrogen's oxygen demand
  • D: counted the moles of products instead of the O2 needed

Problem 10 · FE Environmental, Combustion CO₂ and air

Ethane (C2H6) burns completely with 40% excess air. Taking air as 3.76 mol N2 per mol O2, the moles of air supplied per mole of fuel are most nearly:

  • A 4.90 mol air/mol fuel
  • B 18.4 mol air/mol fuel
  • C 16.7 mol air/mol fuel
  • D 23.3 mol air/mol fuel

Handbook: Thermodynamics, Combustion Processes, FE Reference Handbook 10.6

Show the worked solution
Answer
D (23.3 mol air/mol fuel)
Given
n = 2, excess = 40%, n2 = 3.76, h = 6, two = 2
Find
moles of air per mole of fuel
Handbook
Thermodynamics, Combustion Processes (combustion in air, excess air), page 153
Equation
O2 needed = (3n + 1)/2 per mol CnH2n+2; air = 4.76 × O2 × (1 + excess)
Substitute
  1. C2H6 + 3.5(O2 + 3.76 N2) → 2 CO2 + 3 H2O + 13.16 N2 (theoretical)
  2. Theoretical air = 3.5 × 4.76 = 16.66 mol/mol fuel
  3. Supplied air = 16.66 × (1 + 40/100) = 23.3 mol air/mol fuel
Result
23.3 mol air/mol fuel, 3 significant figures
Check
excess oxygen leaves in the flue gas; the nitrogen passes through unchanged.
Why the others are wrong
  • A: is the oxygen, not the air (nitrogen left out)
  • B: used 3.76 (the nitrogen) instead of 4.76 moles of air per mole of O2
  • C: counted half the hydrogen's oxygen demand

A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

Where is combustion CO₂ and air in the FE Reference Handbook?

Look in the Thermodynamics, Combustion Processes part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.

Are these real FE exam questions?

No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.

How do I check my answer before opening the solution?

Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.

Sources

  1. NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.