A mixture of chromium and zinc weighing 0.362 g was reacted with an excess of hydrochloric acid. After all the metals in the mixture reacted, 225 mL dry of hydrogen gas was collected at \(27^{\circ} \mathrm{C}\) and \(750 .\) torr. Determine the mass percent of \(\mathrm{Zn}\) in the metal sample [Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas; chromium reacts with hydrochloric acid to produce chromium (III) chloride and hydrogen gas. \(]\)

For the reaction of Zinc with Hydrochloric acid: \(Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g) \) For the reaction of Chromium with Hydrochloric acid: \(Cr(s) + 6HCl(aq) \rightarrow CrCl_3(aq) + 3H_2(g) \)

We can use the Ideal Gas Law, \(PV=nRT\), to find the moles (n) of hydrogen gas produced. We are given the volume (V) as 225 mL, which needs to be converted to liters (L), the temperature (T) in Celsius, which needs to be converted to Kelvin (K), and the pressure (P) in torr, which needs to be converted to atm. \(V = 225 \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.225 \mathrm{L} \) \(T = 27^{\circ} \mathrm{C} + 273.15 = 300.15 \mathrm{K} \) \(P = 750 \mathrm{torr} \times \frac{1 \mathrm{atm}}{760 \mathrm{torr}} \approx 0.987 \mathrm{atm} \) The Ideal Gas Law constant, R, is 0.0821 L·atm/mol·K. Now, we can calculate the moles of hydrogen gas produced: \(n = \frac{PV}{RT} = \frac{0.987 \mathrm{atm} \times 0.225 \mathrm{L}}{(0.0821 \mathrm{L\cdot atm/mol\cdot K}) \times 300.15 \mathrm{K}} \approx 0.00903 \mathrm{mol}\)

Let x be the moles of Chromium in the mixture and y be the moles of Zinc in the mixture. Using the stoichiometry of the reactions, we can relate the moles of hydrogen gas produced to the moles of Cr and Zn: \(3x + y = 0.00903 \)

The mass of the mixture (0.362g) is the sum of the mass of Chromium and the mass of Zinc: \(51.996 \cdot x + 65.38 \cdot y = 0.362\)

We need to solve the system of equations: \(3x + y = 0.00903\) \(51.996 \cdot x + 65.38 \cdot y = 0.362\) Using substitution or elimination, we find x ≈ 0.00123 mol and y ≈ 0.00538 mol. Now, we can calculate the mass of Zinc (Zn) in the mixture: Mass of Zn = 65.38g/mol × 0.00538 mol ≈ 0.348g Finally, we can calculate the mass percent of Zn in the mixture: Mass percent of Zn = \(\frac{0.348}{0.362} \times 100\% \approx 96.1\%\) So the mass percent of Zinc in the metal sample is approximately 96.1%.