Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as \(\mathrm{Mn}\) or \(\mathbf{F} \mathrm{e}\)): $$2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)$$ What volume of pure \(\mathrm{O}_{2}(g),\) collected at \(27^{\circ} \mathrm{C}\) and 746 torr, would be generated by decomposition of 125 \(\mathrm{g}\) of a 50.0\(\% \mathrm{by}\) mass hydrogen peroxide solution? Ignore any water vapor that may be present.

The volume of pure \(\mathrm{O}_2(g)\) generated by the decomposition of 125 g of a 50.0% by mass hydrogen peroxide solution, collected at \(27^{\circ} \mathrm{C}\) and 746 torr, is determined in three steps. First, we find the number of moles of \(\mathrm{H}_2\mathrm{O}_2\) in the 50% solution: \(n_{\mathrm{H}_2\mathrm{O}_2} = \frac{0.50 \times 125 \ g}{34.01\ \mathrm{g/mol}}\). Then, using the balanced chemical equation, we find the number of moles of generated \(\mathrm{O}_2\): \(n_{\mathrm{O}_2} = \frac{n_{\mathrm{H}_2\mathrm{O}_2}}{2}\). Finally, we use the Ideal Gas Law to find the volume of \(\mathrm{O}_2\): \(V = \frac{n_{\mathrm{O}_2} \cdot 0.0821 \ \mathrm{atm \cdot L/(mol\cdot K)} \cdot 300\ \mathrm{K}}{\frac{746}{760}\ \mathrm{atm}}\). Calculating these values, we find that the volume of pure \(\mathrm{O}_2(g)\) generated is approximately \(547.2 \ \mathrm{L}\).