Question 14.5: Repeat Example 14.4 if the burning takes place with 80% exce......

The excess air, i.e. 80% or 0.8 will be the surplus air and will reflect as product, hence the chemical reaction will become:

C_3H_8 + 1.8 a(O_2 + 3.78  N_2) \longrightarrow bCO_2 + cH_2O + (0.8a \times d)(3.78  N_2)+ 0.8aO_2

Both side atom conservation gives:
H balance: 2 c = 8 or c = 4.
C balance: b = 3.
O balance: 2 a = z b + c or a = 5.
N balance: 2 × 3.78 a = 2d or d = 18.8
Reaction equation becomes:

C_3H_8 + (9  O_2 + 34.02  N_2) \longrightarrow 3  CO_2 + 4H_2O + 34.02  N_2 + 4  O_2

Mass fraction

for CO_2 = \frac{Mass  of  CO_2}{Mass  of  products} = \frac{Mass  of  CO_2}{Mass  of  CO_2 + Mass  of  H_2O + Mass  of  N_2 + Mass  of  O_2}
= \frac{\left[3 \times 44\right] }{\left[3 \times 44 + 4 \times 18 + 34.02 \times 28 + 4 \times 16\right] } = 0.108

Mass fraction

for H_2O = \frac{Mass  of  H_2O}{Mass  of  products} = \frac{Mass  of  H_2O}{Mass  of  CO_2 + Mass  of  H_2O + Mass  of  N_2 + Mass  of  O_2 }
= \frac{\left[4 \times 18\right] }{\left[3 \times 44 + 4 \times 18 \times 34.02 \times 28 + 4 \times 16 \right] } = 0.059