Step 1. Initial moles of H_{2}O_{2}

$\mathrm{g}\mathrm{total}\mathrm{gas}=150\overline{)\mathrm{mL}}\left(\frac{1.11\mathrm{g}}{1\overline{)\mathrm{mL}}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{g}\mathrm{total}\mathrm{gas}=166.5\mathrm{g}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{moles}{\mathrm{H}}_{2}{\mathrm{O}}_{2}=30\%(166.5\overline{)\mathrm{g}})\left(\frac{1\mathrm{mol}}{34\overline{)\mathrm{g}}}\right)\hspace{0.17em}\phantom{\rule{0ex}{0ex}}\mathrm{moles}{\mathrm{H}}_{2}{\mathrm{O}}_{2}=1.47\mathrm{mol}$

Step 2. Moles of left after 41.3 seconds

$1\mathrm{st}\mathrm{order}\mathrm{reaction}:\phantom{\rule{0ex}{0ex}}\mathrm{ln}{\left[\mathrm{A}\right]}_{\mathrm{t}}=-\mathrm{kt}+\mathrm{ln}\hspace{0.17em}{\left[\mathrm{A}\right]}_{0}\phantom{\rule{0ex}{0ex}}\mathrm{ln}{\left[\mathrm{A}\right]}_{\mathrm{t}}=-\left(0.00752{}\overline{){\mathrm{s}}^{-1}}\right)\left(41.3\overline{){\mathrm{s}}}\right)+\mathrm{ln}\left(1.47\mathrm{mol}\right)\phantom{\rule{0ex}{0ex}}\mathrm{ln}{\left[\mathrm{A}\right]}_{\mathrm{t}}=0.0747\mathrm{mol}\phantom{\rule{0ex}{0ex}}{\left[\mathrm{A}\right]}_{\mathrm{t}}=1.078\mathrm{mol}$

The reaction 2H_{2}O_{2}(*a**q*)→2H_{2}O(*l*)+O_{2}(*g*) is first order in H_{2}O_{2} and under certain conditions has a rate constant of 0.00752 s^{−1} at 20.0 °C. A reaction vessel initially contains 150.0 mL of 30.0% H_{2}O_{2} by mass solution (the density of the solution is 1.11 g/mL). The gaseous oxygen is collected over water at 20.0°C as it forms.

What volume of O_{2} will form in 41.3 seconds at a barometric pressure of 721.1 mmHg. (The vapor pressure of water at this temperature is 17.5 mmHg)

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