Question

Write the balanced reaction equation for the complete combustion of butane, C₄H₁₀, in air. Determine the mass and mole fractions of fuel, oxygen and nitrogen in the reactants. Also, determine the mass and mole fraction of carbon dioxide in the products. Determine the mass and mole air-fuel ratios.

Answer 1

Part a

Balanced equation

C4H10 + 6.5 (O2 + 3.76 N2) = 4CO2 + 5H2O + 24.44N2

Part b

From the stoichiometry of the reaction

Moles of C4H10 = 1

Moles of O2 = 6.5

Moles of N2 = 24.44

Total Moles of reactants = 31.94 moles

Mol fraction of C4H10 = 1/31.94 = 0.0313

Mol fraction of O2 = 6.5/31.94 = 0.2035

Mol fraction of N2 = 24.44/31.94 = 0.7652

Mass of C4H10 = 58 g

Mass of O2 = 32*6.5 = 208 g

Mass of N2 = 24.44*28 = 684.32 g

Total mass = 950.32

Mass fraction of C4H10 = 58/950.32 = 0.0611

Mass fraction of O2 = 208/950.32 = 0.2188

Mass fraction of N2 = 684.32/950.32 = 0.7201

Part c

In the products

Total Moles = 4 mol of CO2 + 5 mol of H2O + 24.44 N2

= 33.44 mol

Mole fraction of CO2 = 4/33.44 = 0.1196

Total mass of products = 4*44 + 5*18 + 24.44*28

= 950.32 g

Mass fraction of CO2 = 4*44/950.32 = 0.1852

Part d

Mass of air = 6.5 x 4.76 mol x 29 g/mol = 897.26 g

Mass of fuel = 1 mol x 58 g/mol = 58 g

Mass of air fuel ratio = 897.26/58 = 15.47 g-air/g-fuel

Moles of air = 6.5*4.76 = 30.94 mol

Moles of fuel = 1 mol

Moles of air-fuel ratio = 30.94/1 = 30.94 mol