91Ó°ÊÓ

The balanced equation for the complete combustion of ethanol is,

$C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)→2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

The balanced equation for the combustion of gasoline is,

$2C_8{H}_{18}\left(l\right)+25O_2\left(g\right)→16C{O}_2\left(g\right)+18H_{2}O\left(l\right)$

The equation for the complete combustion of ethanol is,

$C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)→2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

The volume of ethanol is,

$$\begin{gathered} V=1.00L×\frac{10}{100}×\frac{1000mL}{1L} \\ V=100.0mL \end{gathered}$$

Now, the mass of oxygen gas using the volume of ethanol is,

$$\begin{gathered} =\left(100.0mL\right)×\frac{0.789g C_2{H}_{5}OH}{1mL C_2{H}_{5}OH}×\frac{1mol C_2{H}_{5}OH}{46.0g C_2{H}_{5}OH}×\frac{3mol O_2}{1mol C_2{H}_{5}OH}×\frac{32.0g O_2}{1mol O_2}\backslash \mathrm{hfill} \\ =165.0g \end{gathered}$$

The equation for the combustion of gasoline is,

$2C_8{H}_{18}\left(l\right)+25O_2\left(g\right)→16C{O}_2\left(g\right)+18H_{2}O\left(l\right)$

The volume of gasoline is,

$$\begin{gathered} V=1.00L×\frac{90}{100}×\frac{1000mL}{1L} \\ V=900.0mL \end{gathered}$$

Now, the mass of oxygen gas using the volume of ethanol is,

$$\begin{gathered} =\left(900.0mL\right)×\frac{0.742g C_8{H}_{18}}{1mL C_8{H}_{18}}×\frac{1mol C_8{H}_{18}}{114.0g C_8{H}_{18}}×\frac{12.5mol O_2}{1mol C_8{H}_{18}}×\frac{32.0g O_2}{1mol O_2}\backslash \mathrm{hfill} \\ =2.34×{10}^{3}g \end{gathered}$$

Now, the total mass of the oxygen gas is,

$$\begin{gathered} 165.0g+\left(2.34×{10}^{3}g\right) \\ =2.51×{10}^{3}g \end{gathered}$$

Thus, the mass of the oxygen gas is 2.51x10³ g.