Answer :
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
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