2. (11 marks total) Hydrocarbon fuels have the general chemical formula CH. To i

Question

2. (11 marks total) Hydrocarbon fuels have the general chemical formula CH. To identify an unknown hydrocarbon gas, 0.001 kmol of this fuel are burned and the combustion products are found to have a composition of 13.63% CO, 5.58% H2O, 7.43% O, and 73.36% , BY MASS (note: not the usual volume percent) (a) (4 marks) If the total mass of the products produced from 0.001 kmol of fuel is 0.646 kg, identify the hydrocarbon - i.e. solve for the unknowns n and m, (b) (7 marks) If this reaction is carried out in an adiabatic combustion chamber at constant pressure, the products reach a temperature around 1950 K. Using this information, find the Higher Heating Value (HHV) for this fuel, in kJ/kg. The reactants (fuel and air) enter at 25°C Product Molar mass 0 at 1900K | ? 2 at 2000K fg (kJ/kmol) (kg/kmol) (kJ/kmol) (kJ/kmol) (kJ/kmol) CO2 85429 67613 55434 52551 91450 72689 59199 56141 -393522 H2O (g) 18 32 28 -241827 43 961

Explanation / Answer

CnHm + O2 -> nCO2 + (m/2) H2O

Total Mass = 0.646 Kg

Mass of CO2 = 13.63% = 0.1363 *0.646 Kg = 0.089 Kg or 89 g

Molecular Weight of CO2 = 44 g/mol

Number of moles = Mass / Molecular mass = 2 (n moles of C in the reactant gives n moles of CO2 in the product)

Similarly, Mass of H2O = 0.0558 * 0.646 = 0.036 Kg = 36 g

Molecular Weight of H2O = 18 g/mol

No. of moles = 36 / 18 = 2

No. of moles of H = 2 * no. of moles of H2O = 4 (According to stoichiometry of the reaction, m moles of H in the reactant gives m/2 moles of H2O)

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.