Problem 110
Question
(a) A sample of hydrogen gas is generated in a closed container by reacting \(1.750 \mathrm{~g}\) of zinc metal with \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M}\) hydrochloric acid. Write the balanced equation for the reaction, and calculate the number of moles of hydrogen formed, assuming that the reaction is complete. (b) The volume over the solution in the container is 150 \(\mathrm{mL}\). Calculate the partial pressure of the hydrogen gas in this volume at \(25^{\circ} \mathrm{C}\), ignoring any solubility of the gas in the solution. (c) The Henry's law constant for hydrogen in water at \(25^{\circ} \mathrm{C}\) is \(7.7 \times 10^{-6} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}\). Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part (b)?
Step-by-Step Solution
VerifiedKey Concepts
Limiting Reagent
For example, in the reaction between zinc and hydrochloric acid to produce hydrogen gas, there are two reactants: zinc (\(\text{Zn}\)) and hydrochloric acid (\(\text{HCl}\)).
- To determine the limiting reagent, we calculate moles of each reactant. Here, 1.750 g of zinc equates to about 0.02676 moles.
- Meanwhile, there are 0.0500 moles of hydrochloric acid present.
- The balanced chemical equation for this reaction is \(\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2\), implying that 2 moles of HCl are required for each mole of Zn consumed.
Ideal Gas Law
- \(P\) is the pressure,
- \(V\) is the volume,
- \(n\) represents moles of gas,
- \(R\) is the ideal gas constant (commonly 0.0821 L atm/K mol),
- and \(T\) is the temperature in Kelvin.
- The produced hydrogen gas fills a volume of 0.150 L, and it exists at a temperature of \(25^{\circ} \text{C}\) or 298 K.
- Knowing \(n = 0.0250\) mol, we apply the ideal gas law: \(P = \frac{nRT}{V}\).
Understanding how gases like hydrogen behave under these conditions helps predict outcomes in various chemical processes.
Henry's Law
- In this equation, \(k\) is the Henry's Law constant, differing depending on the gas and the liquid.
- In our exercise, \(k\) is \(7.7 \times 10^{-6}\) mol/m³-Pa for hydrogen in water.
- First, convert the partial pressure from atm to Pa, giving \(P = 414414.25\) Pa.
- This converts to a concentration \(C = 3.19 \times 10^{-3}\) mol/m³, assuming 0.0500 L of water (or 0.0500 m³).
- The dissolved moles of hydrogen are about \(1.60 \times 10^{-4}\) mol.
Henry's Law clarifies the usually small contribution of gas solubility under specific reaction conditions.