Answer
$m_{co_{2}}= 88\text{ kg}\\m_{H20}= 18kg\\m_{O_{2}}=20\text{ kg}\\m_{total}=126\text{ kg}$
Work Step by Step
The stoichiometric combustion equation is
$$
\mathrm{C}_2 \mathrm{H}_2+2.5 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
$$ The combustion equation with $25 \%$ excess oxygen is $$
\mathrm{C}_2 \mathrm{H}_2+(1.25 \times 2.5) \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}+(0.25 \times 2.5) \mathrm{O}_2
$$ or $$
\mathrm{C}_2 \mathrm{H}_2+3.125 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}+0.625 \mathrm{O}_2
$$ The mass of each product and the total mass are $$
\begin{aligned}
m_{\mathrm{CO} 2} & =N_{\mathrm{CO} 2} M_{\mathrm{CO} 2}=(2 \mathrm{kmol})(44 \mathrm{~kg} / \mathrm{kmol})=88 \mathrm{~kg} \\
m_{\mathrm{H} 2 \mathrm{O}} & =N_{\mathrm{H} 2 \mathrm{O}} M_{\mathrm{H} 2 \mathrm{O}}=(1 \mathrm{kmol})(18 \mathrm{~kg} / \mathrm{kmol})=18 \mathrm{~kg} \\
m_{\mathrm{O} 2} & =N_{\mathrm{O} 2} M_{\mathrm{O} 2}=(0.625 \mathrm{kmol})(32 \mathrm{~kg} / \mathrm{kmol})=20 \mathrm{~kg} \\
m_{\text {total }} & =m_{\mathrm{CO} 2}+m_{\mathrm{H} 2 \mathrm{O}}+m_{\mathrm{O} 2}=88+18+20=126 \mathrm{~kg}
\end{aligned}
$$ Then the mass fractions are $$
\begin{aligned}
& \mathrm{mf}_{\mathrm{CO} 2}=\frac{m_{\mathrm{CO} 2}}{m_{\text {total }}}=\frac{88 \mathrm{~kg}}{126 \mathrm{~kg}}=\mathbf{0 . 6 9 8 4} \\
& \mathrm{mf}_{\mathrm{H} 2 \mathrm{O}}=\frac{m_{\mathrm{H} 2 \mathrm{O}}}{m_{\text {total }}}=\frac{18 \mathrm{~kg}}{126 \mathrm{~kg}}=\mathbf{0 . 1 4 2 9} \\
& \mathrm{mf}_{\mathrm{O} 2}=\frac{m_{\mathrm{O} 2}}{m_{\text {total }}}=\frac{20 \mathrm{~kg}}{126 \mathrm{~kg}}=\mathbf{0 . 1 5 8 7}
\end{aligned}
$$ The mass of oxygen per unit mass of fuel burned is determined from $$
\frac{m_{\mathrm{O} 2}}{m_{\mathrm{C} 2 \mathrm{H} 2}}=\frac{(3.125 \times 32) \mathrm{kg}}{(1 \times 26) \mathrm{kg}}=3.846\ \mathrm{kgO}{ }_2 / \mathrm{kgC}_2 \mathrm{H}_2
$$
