Question

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

Short answer

Answer: The approximate percentage of zinc in the Al-Zn alloy is 43.7%.

Step-by-step solution

Find the moles of hydrogen gas produced

First, we need to convert the given temperature and pressure to the appropriate units: Temperature: \(25^{\circ} C = 25+273.15=298.15 K\) Pressure: \(755 mm\,Hg\cdot \frac{1\,atm}{760\,mm \,Hg} \approx 0.994\, atm\) Now we can use the Ideal Gas Law, which states that \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature. We can solve for the number of moles (\(n\)) of hydrogen gas produced: $$ n = \frac{PV}{RT} = \frac{0.994\,atm \times 0.147\,L}{0.0821\,\frac{L \cdot atm}{K \cdot mol} \times 298.15\,K} \approx 0.00601\,mol $$

Calculate the moles of HCl reacted

Now we can use the stoichiometry of the balanced chemical equations to find the moles of HCl reacted. The sum of the stoichiometric coefficients of hydrogen gas produced for both reactions is 2.5: $$ \mathrm{H}^{+} : \mathrm{H}_{2} = 2.5 : 1 $$ So, the moles of HCl reacted are: $$ 0.00601\,mol \times 2.5 = 0.01503\, mol $$

Determine the moles of Al and Zn in the alloy

Let \(x\) be the moles of Al in the alloy and \(y\) the moles of Zn. Based on the mole ratios of Al/HCl and Zn/HCl from the balanced equations: $$ x + 2y = 0.01503 $$

Find the total mass of Al and Zn in the alloy sample

The mass of the alloy is given as 0.2500 g. We can find the total mass of Al and Zn by summing their respective masses in the alloy sample: $$ (26.98\,g/mol)x + (65.38\,g/mol)y = 0.2500\,g $$

Calculate the percentage of zinc in the alloy

We have a system of two linear equations with two variables: $$ \begin{cases} x + 2y = 0.01503 \\ (26.98)x + (65.38)y = 0.2500 \end{cases} $$ Solving this system of equations, we get \(x\approx 0.00663\,mol\) and \(y\approx 0.00420\,mol\). Now we can calculate the percentage of zinc in the alloy: $$ \frac{65.38\,g/mol \times 0.00420\,mol}{0.2500\,g} \times 100\% \approx 43.7\% $$ Therefore, the percentage of zinc in the alloy is approximately 43.7%.