N—butane fuel (c4h10) is burned with a stoichiometric amount of air. determme the mass fraction of each product. also, calculate the mass of carbon dioxide in the products and mass of air required when 6.14 kg of fuel is burned.

The balanced equation for the burning of N-butane is;
                      
                          2C₄H₁₀  +  13O₂  →   8CO₂   +  10H₂O
stoi. ratio               2    :         13       :      8        :      10
molar mass          58              32               44           18
moles                    X             13 X/2        8X/2          10X/2
                            105.86      688.10      423.45       529.31
produced mass                                    18632.8 g    9527.6 g

moles of n-butane = X = 6.14 x 10³ g / 58 g mol⁻¹ 
hence other moles of compounds can be calculated.

Mass = molar mass * moles

mass fraction = mass / total mass

hence the mass fraction of
    CO₂ = 18632.8 / 6.14 x 10³  = 3.03
    H₂O = 9527.6 / 6.14 x 10³    = 1.55

the required mass of air = 688.10 x 32 = 22019.2 g = 2.20 kg

The answer is b i think