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(a) A sample of hydrogen gas is generated in a closed container by reacting \(2.050 \mathrm{~g}\) of zinc metal with \(15.0 \mathrm{~mL}\) of \(1.00 \mathrm{M}\) sulfuric 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 is \(122 \mathrm{~mL}\). 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 \(7.8 \times 10^{-4} \mathrm{~mol} / \mathrm{L}-\mathrm{atm} .\) 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)?

Short Answer

Expert verified
The balanced chemical equation for the reaction between zinc (Zn) and sulfuric acid (H鈧係O鈧) is: Zn + H鈧係O鈧 -> ZnSO鈧 + H鈧. There are 0.0314 mol of hydrogen gas (H鈧) formed. The partial pressure of H鈧 gas above the solution at 25掳C is 6.256 atm. The number of moles of H鈧 dissolved in the solution is 7.319 脳 10鈦烩伒 mol, and the fraction of dissolved gas molecules is approximately 0.23%. Since the dissolved fraction is small, it was reasonable to ignore dissolved hydrogen in part (b).

Step by step solution

01

Write the balanced chemical equation for the reaction between zinc and sulfuric acid

The balanced chemical equation for the reaction between zinc (Zn) and sulfuric acid (H鈧係O鈧) can be written as: Zn + H鈧係O鈧 -> ZnSO鈧 + H鈧 Now, we can proceed to calculate the number of moles of hydrogen gas formed.
02

Calculate the number of moles of hydrogen formed

Given that there are 2.050 g of zinc (Zn), we can first calculate the number of moles of Zn. The molar mass of Zn is 65.38 g/mol, so the number of moles of Zn is: 2.050 g / 65.38 g/mol = 0.0314 mol Since the balanced equation shows a 1:1 molar ratio between Zn and H2, there will also be 0.0314 mol of hydrogen gas (H鈧) formed.
03

Calculate the partial pressure of hydrogen gas

We need to calculate the partial pressure of hydrogen gas above the solution at 25掳C, ignoring any solubility of the gas in the solution. Since we only have H鈧 gas in the 122 mL container, we can obtain the pressure using the ideal gas law: PV=nRT where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin. First, we need to convert the volume from mL to L and the temperature from Celsius to Kelvin: V = 122 mL = 0.122 L T = 25掳C + 273.15 = 298.15 K The ideal gas constant, R, can be taken as 0.08206 L路atm/mol路K. Now we can solve for the pressure (P): P = nRT / V = (0.0314 mol)(0.08206 L路atm/mol路K)(298.15 K) / 0.122 L = 6.256 atm
04

Estimate the number of moles of hydrogen gas that remain dissolved in the solution

The Henry's law constant (K_H) for hydrogen in water at 25掳C is 7.8 脳 10鈦烩伌 mol / L路atm, which relates the solubility of hydrogen gas in the solution (C_H鈧) to the partial pressure of hydrogen gas (P_H鈧): C_H鈧 = K_H 脳 P_H鈧 Therefore, we can calculate the solubility of hydrogen in the solution: C_H鈧 = (7.8 脳 10鈦烩伌 mol / L路atm) 脳 6.256 atm = 4.879 脳 10鈦宦 mol / L Now, multiply by the volume (15.0 mL = 0.015 L) to get the number of moles of H鈧 dissolved in the solution: 4.879 脳 10鈦宦 mol / L 脳 0.015 L = 7.319 脳 10鈦烩伒 mol
05

Calculate the fraction of dissolved gas molecules and determine whether it is reasonable to ignore dissolved hydrogen

To calculate the fraction of dissolved gas molecules, divide the number of moles of dissolved H鈧 by the total number of moles of H鈧: fraction = (7.319 脳 10鈦烩伒 mol) / (0.0314 mol) = 0.00233 Since the fraction of dissolved hydrogen is small (approx. 0.23%), it was reasonable to ignore any dissolved hydrogen in part (b).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reaction Equation Balancing
Understanding the stoichiometry of gas reactions hinges on the ability to balance chemical equations. A balanced chemical equation accurately represents the conservation of mass and the mole ratios of reactants and products involved in a chemical reaction. During the reaction between zinc and sulfuric acid to generate hydrogen gas, a balanced equation is crucial for predicting the stoichiometry and determining the mole ratio between the reactants and the products.

For instance, the equation Zn + H2SO4 -> ZnSO4 + H2 shows a 1:1 ratio, signaling that one mole of zinc reacts with one mole of sulfuric acid to produce one mole of hydrogen gas. In educational content, it's central to highlight that each atom's count must be the same on both sides of the equation, which is a fundamental rule in the balancing process.
Ideal Gas Law Application
The application of the ideal gas law is essential in calculating the properties of gases under various conditions. In the example of hydrogen gas generated from the reaction of zinc metal with sulfuric acid, the ideal gas law assists in finding the partial pressure of the gas.

The ideal gas law, represented by the equation PV=nRT, defines the relationship between pressure (P), volume (V), temperature (T), and the number of moles (n) of an ideal gas, where R is the universal gas constant. To solve for pressure, as the exercise requires, one must ensure that the volume is in liters, temperature is in Kelvin, and the gas constant's value corresponds to the units of the other terms. Practical educational content will explain each step clearly and emphasize the importance of unit consistency to yield accurate results.
Henry's Law and Gas Solubility
Henry's law describes the solubility of a gas in a liquid at a constant temperature, stating that the amount of gas dissolved is directly proportional to its partial pressure above the liquid. The law is captured by the equation Cg = kH * Pg, where Cg is the concentration of the gas in the liquid, kH is Henry's law constant, and Pg is the partial pressure of the gas.

In the problem of hydrogen in water, Henry's law is used to estimate the number of hydrogen molecules dissolved. This concept is pivotal in scenarios where the solubility of gases plays a crucial role, such as in environmental studies of gas exchange between the atmosphere and oceans, or in designing carbonated beverages in the food industry. For students, explicating how to apply Henry's law and the implications of assuming its validity enhances the comprehension and significance of this concept in real-world scenarios.

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Most popular questions from this chapter

Describe how you would prepare each of the following aqueous solutions: (a) \(1.50\) L of \(0.110 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) solution, starting with solid \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} ;\) (b) \(120 \mathrm{~g}\) of a solution that is \(0.65 \mathrm{~m}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), starting with the solid solute; (c) \(1.20 \mathrm{~L}\) of a solution that is \(15.0 \% \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass (the density of the solution is \(1.16 \mathrm{~g} / \mathrm{mL}\) ), starting with solid solute; (d) a \(0.50 \mathrm{M}\) solution of \(\mathrm{HCl}\) that would just neutralize \(5.5 \mathrm{~g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) starting with \(6.0\) M \(\mathrm{HCl}\).

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