Answer
$M_{m}= 27.84\text{ kg/kmol}$
Work Step by Step
The reaction with stoichiometric air is $$
\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+a_{s h}\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}+a_{t h} \times 3.76 \mathrm{~N}_2
$$ where $$
0.5+a_{t h}=2+1.5 \longrightarrow a_{t h}=3
$$ Substituting, $$
\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+3\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}+3 \times 3.76 \mathrm{~N}_2
$$ The balanced reaction equation with incomplete combustion is $$
\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+3\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 2\left(0.90 \mathrm{CO}_2+0.05 \mathrm{CO}\right)+3\left(0.95 \mathrm{H}_2 \mathrm{O}+0.1 \mathrm{OH}\right)+b \mathrm{O}_2+3 \times 3.76 \mathrm{~N}_2
$$ $\mathrm{O}_2$ balance: $\quad 0.5+3=1.8+0.05+3.15 / 2+b \rightarrow \quad b=0.075$
which can be written as $$
\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+3\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 1.8 \mathrm{CO}_2+0.1 \mathrm{CO}+2.85 \mathrm{H}_2 \mathrm{O}+0.3 \mathrm{OH}+0.075 \mathrm{O}_2+11.28 \mathrm{~N}_2
$$ The total moles of the products is $$
N_m=1.8+0.1+2.85+0.3+0.075+11.28=16.41\ \mathrm{kmol}
$$ The apparent molecular weight of the product gas is $$
M_m=\frac{m_m}{N_m}=\frac{(1.8 \times 44+0.1 \times 28+2.85 \times 18+0.3 \times 17+0.075 \times 32+11.28 \times 28) \mathrm{kg}}{16.41 \mathrm{kmol}}=\\\mathbf{2 7 . 8 4}\ \mathrm{kg} / \mathrm{kmol}
$$
