Consider the unbalanced chemical equation \(\mathrm{Cl}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HClO}_{4}\) The reaction is carried out at \(82.0 \%\) yield and gives \(52.8 \mathrm{~g}\) of \(\mathrm{HClO}_{4}\) (a) What is the theoretical yield of \(\mathrm{HClO}_{4}\) ? (b) How many grams of \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) and \(\mathrm{H}_{2} \mathrm{O}\) were consumed in the reaction?

To balance the given chemical equation, we need to adjust the coefficients of the reactants and products so that atoms of each element are equal on both sides of the arrow. The balanced chemical equation for the given reaction is: \[\mathrm{Cl}_{2}\mathrm{O}_{7} + 2\mathrm{H}_{2} \mathrm{O} \rightarrow 2\mathrm{HClO}_{4}\]

Given that actual yield is \(82.0 \% \) which produced \(52.8 \mathrm{~g}\) of \(\mathrm{HClO}_{4}\). We need to find out the theoretical yield. The formula for finding the yield percent is: \[Actual \ Yield \ = \% \ Yield \times Theoretical \ Yield\] Now, let's rearrange the formula to find the theoretical yield. \[Theoretical \ Yield = \frac{Actual \ Yield}{\% \ Yield}\] By plugging the given numbers into the formula, we have: \(Theoretical \ Yield = \frac{52.8 \mathrm{~g}}{0.820}\) \(Theoretical \ Yield = 64.390243902 \mathrm{~g} \approx 64.4 \mathrm{~g}\) So, the theoretical yield of \(\mathrm{HClO}_{4}\) is approximately \(64.4 \mathrm{~g}\).

Now that we have the theoretical yield of \(\mathrm{HClO}_{4}\), we can determine the quantity of consumed \(\mathrm{Cl}_2\mathrm{O}_7\) and \(\mathrm{H}_2\mathrm{O}\) by using the stoichiometry method. First, let's find the mole ratios between \(\mathrm{HClO}_{4}\), \(\mathrm{Cl}_2\mathrm{O}_7\), and \(\mathrm{H}_2\mathrm{O}\) obtained from the balanced chemical equation: \[\frac{2\, moles\,of\, \mathrm{HClO}_{4}}{1\, mole\, of\, \mathrm{Cl}_2\mathrm{O}_7} = \frac{2\, moles\, of\, \mathrm{HClO}_{4}}{2\, moles\, of\, \mathrm{H}_2\mathrm{O}}\] Now, let's find the moles of \(\mathrm{HClO}_{4}\) from the obtained theoretical yield: \[moles\,of\,\mathrm{HClO}_{4} = \frac{64.4\,\mathrm{g}}{100.46\,\frac{\mathrm{g}}{\mathrm{mol}}} = 0.641\,\mathrm{mol}\] Using the mole ratios, we can determine the moles of \(\mathrm{Cl}_2\mathrm{O}_7\) and \(\mathrm{H}_2\mathrm{O}\) consumed in the reaction: \[moles\, of\, \mathrm{Cl}_2\mathrm{O}_7 = 0.641\,\mathrm{mol}\, of\, \mathrm{HClO}_{4} \times \frac{1\, mole\, of\, \mathrm{Cl}_2\mathrm{O}_7}{2\, moles\, of\, \mathrm{HClO}_{4}} = 0.3205\,\mathrm{mol}\] \[moles\, of\, \mathrm{H}_2\mathrm{O} = 0.641\,\mathrm{mol}\, of\, \mathrm{HClO}_{4} \times \frac{2\, moles\, of\, \mathrm{H}_2\mathrm{O}}{2\, moles\, of\, \mathrm{HClO}_{4}} = 0.641\,\mathrm{mol}\] Finally, let's convert moles of \(\mathrm{Cl}_2\mathrm{O}_7\) and \(\mathrm{H}_2\mathrm{O}\) back into grams: \[grams\, of\, \mathrm{Cl}_2\mathrm{O}_7 = 0.3205\,\mathrm{mol} \times 182.903\,\frac{\mathrm{g}}{\mathrm{mol}} = 58.6\,\mathrm{g}\] \[grams\, of\, \mathrm{H}_2\mathrm{O} = 0.641\,\mathrm{mol} \times 18.0153\,\frac{\mathrm{g}}{\mathrm{mol}} = 11.542\,\mathrm{g} \approx 11.5\,\mathrm{g}\] So, about \(58.6\,\mathrm{g}\) of \(\mathrm{Cl}_2\mathrm{O}_7\) and \(11.5\,\mathrm{g}\) of \(\mathrm{H}_2\mathrm{O}\) were consumed in the reaction.