Analysis of an Aluminum-Zinc Alloy Trial [ Mass gelatin capsule 2134 3218 1542 406 723 Mass of alloy sample plus capsule Mass empty beaker Mass of beaker plus displaced water Atmospheric pressure Temperature Calculations Mass alloy sample Mass of displaced water Volume of displaced water (density of water: 1.00 g/mL) Volume of H2 gas Temperature of H2 Vapor pressure of water at room temperature: 17.5 mmHg (Appendix I) n Hg Pressure of dry H2 gas Hg: AlTi MOK > Moles H2 in sample, "it; Moles H2 per gram of sample. Nil; mokvg Al (read from graph) Unknown
Given Data:
- Mass of gelatin capsule: (2134, 3218, 1542, 406, 723) – likely trial masses; we’ll select one value as an example.
- Mass of alloy sample plus capsule: (value not explicitly separated)
- Mass of empty beaker: not specified
- Mass of beaker plus displaced water: not specified
- Atmospheric pressure: not specified; assume 760 mmHg (standard)
- Temperature: not specified; assume 25°C
- Vapor pressure of water at 25°C: 17.5 mmHg
- Density of water: 1.00 g/mL
Let’s assume and reconstruct plausible values to perform an example calculation.
Example Data (Assumed from structure):
- Mass of alloy + capsule: 3.218 g
- Mass of capsule: 2.134 g
→ Mass of alloy sample: 3.218 g – 2.134 g = 1.084 g - Mass of displaced water: 154.2 g
→ Volume of water displaced = 154.2 mL = volume of H₂ gas collected - Temperature of gas = 25°C = 298 K
- Atmospheric pressure: 760 mmHg
- Vapor pressure of water: 17.5 mmHg
→ Pressure of dry H₂ gas: 760 – 17.5 = 742.5 mmHg = 0.977 atm
1. Moles of H₂ Gas (Ideal Gas Law):
Use:
PV=nRTPV = nRTPV=nRT
n=PVRTn = \frac{PV}{RT}n=RTPV
- P: 0.977 atm
- V: 154.2 mL = 0.1542 L
- R: 0.0821 L·atm/mol·K
- T: 298 K
n=(0.977)(0.1542)(0.0821)(298)≈0.150624.486≈0.00615 mol H2n = \frac{(0.977)(0.1542)}{(0.0821)(298)} \approx \frac{0.1506}{24.486} \approx 0.00615 \text{ mol H}_2n=(0.0821)(298)(0.977)(0.1542)≈24.4860.1506≈0.00615 mol H2
2. Moles of H₂ per gram of sample:
0.00615 mol1.084 g=0.00567 mol/g\frac{0.00615 \text{ mol}}{1.084 \text{ g}} = 0.00567 \text{ mol/g}1.084 g0.00615 mol=0.00567 mol/g
3. Determine % Aluminum Using Calibration Graph:
Let’s assume a calibration graph indicates moles H₂/g for pure Al is 0.037 and pure Zn is 0 (Zn does not produce H₂ with acid in this context). Then: %Al=0.005670.037×100≈15.3%\% \text{Al} = \frac{0.00567}{0.037} \times 100 \approx 15.3\%%Al=0.0370.00567×100≈15.3%
Conclusion
This experiment determines the composition of an aluminum-zinc alloy by analyzing the volume of hydrogen gas produced when the alloy reacts with hydrochloric acid. Aluminum reacts with HCl to release hydrogen gas: 2Al+6HCl→2AlCl3+3H22Al + 6HCl \rightarrow 2AlCl_3 + 3H_22Al+6HCl→2AlCl3+3H2
Zinc may also react with HCl, but in some versions of this experiment, it is assumed that aluminum is the primary H₂ source due to conditions or prior calibration. The volume of hydrogen gas collected over water reflects the aluminum content. However, because water vapor also exerts pressure, we must correct the pressure of the collected gas by subtracting the vapor pressure of water at the experiment temperature (17.5 mmHg at 25°C).
Using the Ideal Gas Law, the number of moles of hydrogen gas produced is calculated. This value, divided by the mass of the alloy sample, gives the number of moles of hydrogen per gram of sample. This value is then compared to standard values obtained from a calibration graph—pure aluminum produces a known maximum amount of hydrogen, while zinc does not. The proportion of hydrogen gas produced per gram directly correlates with the percentage of aluminum in the sample.
In our calculation, the sample produced 0.00615 mol H₂ from 1.084 g alloy, equating to 0.00567 mol/g. Comparing this to the 0.037 mol/g standard for pure aluminum, we estimate that the alloy is approximately 15.3% aluminum by mass. The rest of the alloy is presumably zinc, completing the analysis.
