A \(0.2500-\mathrm{g}\) sample of an \(\mathrm{Al}-\mathrm{Zn}\) alloy reacts with \(\mathrm{HCl}\) to form hydro- gen gas: $$ \begin{aligned} &\mathrm{Al}(s)+3 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+\frac{3}{2} \mathrm{H}_{2}(g) \\ &\mathrm{Zn}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{H}_{2}(g) \end{aligned} $$ The hydrogen produced has a volume of \(0.147 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\) and \(755 \mathrm{~mm} \mathrm{Hg}\) What is the percentage of zinc in the alloy?

The Ideal Gas Law relates the pressure, volume, temperature, and amount of a gas in moles: PV = nRT First, we need to convert the given values to the appropriate units: - Pressure (P): Convert 755 mmHg to atm by dividing by 760: \(P=\frac{755}{760} \ \text{atm}\) - Volume (V): Given as 0.147 L - Temperature (T): Convert 25°C to Kelvin by adding 273.15: \(T=25+273.15=298.15 \ \text{K}\) Now, we can solve for the amount of hydrogen gas produced (n) using the Ideal Gas Law and the gas constant R = 0.0821 L atm/(mol K): $$n=\frac{PV}{RT}=\frac{(\frac{755}{760}\ \text{atm})(0.147 \ \text{L})}{(0.0821\ \frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}})(298.15 \ \text{K})}= 0.00601503 \ \text{mol}$$

We know that the total moles of hydrogen gas produced comes from both the Al and Zn reactions. Since 3 moles of H2 are produced for every mole of Al and 1 mole of H2 is produced for every mole of Zn, we can set up a system of equations to determine the moles of Al and Zn: $$x + y = 0.00601503 \ \text{mol H2} \ (1)$$ $$\frac{3}{2}x + y = 0.00601503 \ \text{mol H2} \ (2)$$ Where x is the moles of Al and y is the moles of Zn. To solve the system of equations, we'll subtract equation (1) from equation (2) to eliminate y: $$\frac{1}{2}x = 0.003007515 \ \text{mol}$$ Then, we can solve for x and y: $$x = 2(0.003007515 \ \text{mol}) = 0.00601503 \ \text{mol Al}$$ $$y = 0.00601503 - 0.00601503 = 0 \ \text{mol Zn}$$

Now we have the moles of Al and Zn in the sample. We can convert these values to mass using their respective molar masses (Al: 26.98 g/mol and Zn: 65.38 g/mol): $$\text{mass of Al}= (0.00601503 \ \text{mol})(26.98 \ \frac{\text{g}}{\text{mol}})= 0.1622 \ \text{g Al} $$ $$\text{mass of Zn}= (0 \ \text{mol})(65.38 \ \frac{\text{g}}{\text{mol}})= 0 \ \text{g Zn} $$

Finally, we can calculate the percentage of zinc in the sample using the mass of Zn and the total mass of the sample: $$\text{percentage of Zn}=\frac{\text{mass of Zn}}{\text{total mass of sample}} \times 100=\frac{0 \ \text{g Zn}}{0.25 \ \text{g sample}} \times 100=0 \%$$ The percentage of zinc in the alloy is 0%.