A mixture of chromium and zinc weighing \(0.362 \mathrm{g}\) was reacted with an excess of hydrochloric acid. After all the metals in the mixture reacted, \(225 \mathrm{mL}\) 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.]

To calculate the moles of hydrogen gas produced, we can use the ideal gas equation: \(PV = nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, we need to convert the pressure to atmospheres and the temperature to Kelvin: Pressure (atm) = \(\frac{750 \text{ torr}}{760\text{ torr/atm}}\) = 0.9868 atm Temperature (K) = 27°C + 273.15 = 300.15 K Volume (L) = \(\frac{225 \text{ mL}}{1000\text{ mL/L}}\) = 0.225 L Now we can calculate the number of moles of hydrogen gas produced: \(n = \frac{PV}{RT}\) \(n_{\text{H}_{2}} = \frac{(0.9868\text{ atm})(0.225\text{ L})}{(0.0821\text{ L atm/mol K})(300.15\text{ K})}\) \(n_{\text{H}_{2}} = 0.00905 \text{ mol}\)

Now we will use the stoichiometry of the reactions to determine the moles of zinc and chromium reacted. The balanced chemical equations are: \(Zn + 2HCl \rightarrow ZnCl_{2} + \text{H}_{2}\) \(2Cr + 6HCl \rightarrow 2CrCl_{3} + 3\text{H}_{2}\) From these equations, we see that 1 mole of zinc will produce 1 mole of hydrogen gas, and 1 mole of chromium will produce 1.5 moles of hydrogen gas. We can use these ratios to set up a system of equations: \(n_{Zn} + n_{Cr} = n_{\text{H}_{2}}\) \(\frac{1}{2} n_{Cr} = n_{Zn}\) We can substitute the value of \(n_{\text{H}_{2}}\) into the first equation: \(n_{Zn} + n_{Cr} = 0.00905 \text{ mol}\) And substitute the value of \(\frac{1}{2} n_{Cr}\) for \(n_{Zn}\) in the second equation: \(\frac{1}{2} n_{Cr} + n_{Cr} = 0.00905 \text{ mol}\)

First, we will find the moles of chromium: \(1.5 n_{Cr} = 0.00905 \text{ mol}\) \(n_{Cr} = 0.00603 \text{ mol}\) Now, we can use the value of \(\frac{1}{2} n_{Cr}\) to find the moles of zinc: \(n_{Zn} = \frac{1}{2} n_{Cr} = \frac{1}{2} (0.00603 \text{ mol}) = 0.00302 \text{ mol}\) Next, we can calculate the mass of zinc and chromium in the sample: Mass of zinc = \(m_{Zn} = n_{Zn} \times M_{Zn} = 0.00302 \text{ mol} \times 65.38\text{ g/mol} = 0.197 \text{ g}\) Mass of chromium = \(m_{Cr} = n_{Cr} \times M_{Cr} = 0.00603 \text{ mol} \times 51.996\text{ g/mol} = 0.313 \text{ g}\) Finally, we can calculate the mass percent of zinc in the sample: Mass percent of zinc = \(\frac{m_{Zn}}{m_{Zn} + m_{Cr}} \times 100\% = \frac{0.197\text{ g}}{0.197\text{ g} + 0.313\text{ g}} \times 100\% = 38.7\%\) So, the mass percent of zinc in the metal sample is 38.7%.