Question

2C₁₂H₂₆(l) + 37O₂(l) -> 24CO₂(g) + 26 H₂O(g)

Calculate the volume of the gas produced in the rocket engine by burning 1 kg of C₁₂H₂₆ at 1 atm, assuming that all gas has a temperature of 3000^oC (assume that CO₂ and H₂O gasses are ideal)

Answer 1

2C12H26(l) + 37O2(l) -> 24CO2(g) + 26 H2O(g)

Moles of C12H26 = mass/molecular weight

= 1000g / (12*12+26*1)g/mol

= 5.882 mol

From the stoichiometry of the reaction

2 mol C12H26 produces = 24 mol CO2

5.882 mol C12H26 produces = 24*5.882/2 = 70.588 mol CO2

Moles of H2O produced

= 26 mol H2O x 5.882 mol C12H26 / 2 mol C12H26

= 76.47 mol

Total Moles produced = 76.47 + 70.588

= 147.058 mol

From the ideal gas equation

Volume of ideal gas = nRT/P

= 147.058 mol x 0.0821 L-atm/mol-K x (3000+273)K / 1atm

V = 39516.59 L

Or

= 39516.59 L x 1m3/1000L

V = 39.516 m3