Problem 90 - Chemistry: Principles and Reactions

Question

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?

Step-by-Step Solution

Answer

Answer: The percentage of zinc in the Al-Zn alloy sample is 0%.

Step 1: Convert the hydrogen gas volume to moles using the Ideal Gas Law

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): $$n=\frac{PV}{RT}=\frac{(\frac{755}{760}\ \text{atm})(0.147 \ \text{L})}{(0.0821\ \frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}})(298.15 \ \text{K})}= 0.00601503 \ \text{mol}$$

Step 2: Calculate the moles of Al and Zn in the sample

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: $$x = 2(0.003007515 \ \text{mol}) = 0.00601503 \ \text{mol Al}$$ $$y = 0.00601503 - 0.00601503 = 0 \ \text{mol Zn}$$

Step 3: Calculate the mass of Al and Zn in the sample

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): $$\text{mass of Al}= (0.00601503 \ \text{mol})(26.98 \ \frac{\text{g}}{\text{mol}})= 0.1622 \ \text{g Al} $$ $$\text{mass of Zn}= (0 \ \text{mol})(65.38 \ \frac{\text{g}}{\text{mol}})= 0 \ \text{g Zn} $$

Step 4: Calculate the percentage of zinc in the sample

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%.

Key Concepts

Gas Reactions

When dealing with gas reactions, understanding how gases interact is crucial. In our exercise, solid metals react with hydrochloric acid to produce hydrogen gas. Let's break it down:

• Aluminum (Al) and zinc (Zn) are the reactive metals in our alloy.
• When these metals react with hydrogen ions ( H^+ ), they produce metal ions and hydrogen gas ( H_2 ).

This transformation is a classic example of gas reactions where a solid substrate interacts with a reactive solution to evolve gas. Remember:

• Aluminum generates hydrogen gas as 3 moles of hydrogen (H_2) for every 2 moles of aluminum ( Al ).
• Zinc produces 1 mole of H_2 for each mole of Zn .

This setup helps us interpret chemical reactions that release gases and illustrates how these reactions tie into broader chemistry concepts, like the use of stoichiometry to quantify reaction components.

Stoichiometry Calculations

Stoichiometry is your chemistry toolkit for measuring and calculating amounts in chemical reactions. In our problem, we calculated the moles of reactants responsible for producing hydrogen gas. A step-by-step breakdown:

• Use the Ideal Gas Law ( PV=nRT ) to find the number of moles of hydrogen gas produced under specific conditions.
• Translate the reaction equations into stoichiometric relationships. (e.g. Al gives 3/2 mol of H_2 while Zn gives 1 mol of H_2)
• Simplify the problem into a system of equations to separate the contributions of aluminum and zinc to hydrogen production.
• Solve the equations to identify moles of individual metals reacting.

This technique is essential for interpreting the products of chemical reactions accurately. It ensures a balanced approach in quantitative analysis, providing a clear picture of how each component contributes to the overall reaction outcome.

Alloy Composition Analysis

The composition of an alloy—the mix of its constituent components—determines its attributes and application. With this analysis, you can ascertain the precise makeup of the metals involved. Considerations for analyzing an alloy like Al-Zn :

• Identify the mass components once you know their moles ( using molar masses: Al is ~26.98 g/mol, Zn is ~65.38 g/mol).
• Calculate the mass percentage of each component. In our exercise, aluminum's mass was crucial because zinc was delineated to contribute no mass (hence, 0% Zn).
• Provide the composition as a percentage which gives an easy way to gauge the relative amount of each metal in the sample.

Such analysis aids in alloy refinement and optimization for specific uses. By understanding the component ratios, one can predict how the alloy will perform under various conditions, a critical insight for materials science and engineering.