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In a "Kipp generator", hydrogen gas is produced when zinc flakes react with hydrochloric acid: $$2 \mathrm{HCl}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ If \(30.0 \mathrm{~mL}\) of wet \(\mathrm{H}_{2}\) is collected over water at \(20^{\circ} \mathrm{C}\) and a barometric pressure of \(101.33 \mathrm{kPa}\), how many grams of \(\mathrm{Zn}\) have been consumed? (The vapor pressure of water is tabulated in Appendix B.)

Short Answer

Expert verified
Approximately 0.0802 grams of zinc have been consumed in the Kipp generator.

Step by step solution

01

Determine the partial pressure of hydrogen gas

To determine the amount of hydrogen gas produced, first find the partial pressure of hydrogen gas. Subtract the vapor pressure of water at 20°C from the given barometric pressure. Partial Pressure of H₂ (P_H₂) = Barometric Pressure - Vapor Pressure of Water At 20°C, the vapor pressure of water can be found using the tabulated data in Appendix B, which states that it's 2.34 kPa. So: P_H₂ = 101.33 kPa - 2.34 kPa P_H₂ = 98.99 kPa
02

Calculate the moles of hydrogen gas produced

Now that we have the partial pressure of hydrogen gas, we can use the Ideal Gas Law to find the moles of hydrogen gas produced. The Ideal Gas Law is given by: PV = nRT Where P is the pressure (in kPa), V is the volume (in L), n is the number of moles, R is the ideal gas constant (8.314 J/(mol·K)), and T is the temperature (in K). We need to convert the given volume and temperature to appropriate units. Volume of H₂: 30.0 mL = 0.030 L Temperature: 20°C = 293.15 K Rearrange the Ideal Gas Law to solve for n: n = PV/RT Plug in the values: n = (98.99 kPa)(0.030 L) / (8.314 J/(mol·K))(293.15 K) n = 0.001228 mol We have approximately 0.001228 moles of hydrogen gas.
03

Determine the amount of zinc consumed

Now we will use the stoichiometry of the reaction to determine the amount of zinc consumed. The balanced chemical equation is: 2 HCl(aq) + Zn(s) → ZnCl₂(aq) + H₂(g) From the stoichiometry, we can see that 1 mole of H₂ is produced for 1 mole of Zn consumed. Therefore, the moles of Zn consumed are equal to the moles of H₂ produced. Moles of Zn consumed = Moles of H₂ produced = 0.001228 mol
04

Convert moles of zinc to grams

Finally, convert the moles of zinc consumed to grams using the molar mass of zinc, which is approximately 65.38 g/mol. Grams of Zn consumed = Moles of Zn consumed × Molar mass of Zn Grams of Zn consumed = 0.001228 mol × 65.38 g/mol Grams of Zn consumed ≈ 0.0802 g Therefore, approximately 0.0802 grams of zinc have been consumed.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation that describes the behavior of gases. It's expressed as \( PV = nRT \). This equation relates four key variables:
  • \( P \) - Pressure of the gas
  • \( V \) - Volume of the gas
  • \( n \) - Number of moles of the gas
  • \( R \) - Ideal Gas Constant (8.314 J/(mol·K))
  • \( T \) - Temperature in Kelvin
To solve for the number of moles \( n \), we rearrange the equation to \( n = \frac{PV}{RT} \).
In our exercise, the pressure \( P \) is adjusted for the partial pressure of hydrogen after accounting for water vapor pressure. The volume \( V \) is given in liters, and the temperature \( T \) is converted to Kelvin. These conversions are essential to accurately use the Ideal Gas Law for calculations.
Zinc Reaction
The reaction between zinc and hydrochloric acid is quite fascinating. In this process, zinc flakes react with hydrochloric acid to produce zinc chloride and hydrogen gas.
This is a single replacement reaction where zinc, being more reactive, displaces the hydrogen in hydrochloric acid, forming zinc chloride in solution and releasing hydrogen gas. It's an example of the reactivity series, which predicts the ability of an element to displace another in a chemical reaction.
The balanced equation for this reaction is:
\[ \text{2 HCl(aq) + Zn(s) } \rightarrow \text{ ZnCl}_2\text{(aq) + H}_2\text{(g)} \]This equation shows that one mole of zinc reacts with two moles of hydrochloric acid to produce one mole of hydrogen gas.
Hydrogen Gas Production
Hydrogen gas production in chemical reactions like the one with zinc is a well-recorded process. As zinc reacts with hydrochloric acid, we observe bubbling due to hydrogen gas being released.
  • Hydrogen is collected over water, meaning it is gathered as a wet gas with water vapor mixed in.
  • To find out how much pure hydrogen gas is produced, you consider the pressure exerted by the water vapor, known as the vapor pressure.
  • For accurate calculations, you subtract the water vapor pressure from the total barometric pressure to get the partial pressure of the hydrogen.
This allows us to compute the exact amount of hydrogen produced using the Ideal Gas Law once the adjustments are made for vapor pressure.
Chemical Equations
Chemical equations are concise representations of chemical reactions. They list reactants and products, their phases (solid, liquid, aqueous, gas), and the stoichiometric ratios involved.
The balanced chemical equation in our problem is crucial as it reflects the exact proportions of reactants needed and products formed:
\[ \text{2 HCl(aq) + Zn(s) } \rightarrow \text{ ZnCl}_2\text{(aq) + H}_2\text{(g)} \]
  • "Balanced" means the number of atoms for each element is the same on both sides of the equation.
  • Stoichiometry, which is the calculation of reactants and products in chemical reactions, is used alongside the balanced equation to interpret chemical changes.
This is essential for determining how much zinc is consumed for a given amount of hydrogen gas produced, offering insights into the conversion of reactants to products.

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Most popular questions from this chapter

In the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below \(100^{\circ} \mathrm{C}\) in a boiling-water bath and determine the mass of vapor required to fill the bulb. From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, \(1.012 \mathrm{~g}\); volume of bulb, \(354 \mathrm{~cm}^{3}\); pressure, \(98.93 \mathrm{kPa} ;\) temperature, \(99{ }^{\circ} \mathrm{C}\).

Which one or more of the following statements are true? (a) \(\mathrm{O}_{2}\) will effuse faster than \(\mathrm{Cl}_{2}\). (b) Effusion and diffusion are different names for the same process. (c) Perfume molecules travel to your nose by the process of effusion. (d) The higher the density of a gas, the shorter the mean free path.

You have a sample of gas at \(0^{\circ} \mathrm{C}\). You wish to increase the \(\mathrm{rms}\) speed by a factor of \(3 .\) To what temperature should the gas be heated?

Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{~N}_{2}, 15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2},\) and \(6.2 \%\) water vapor. (a) If the total pressure of the gases is \(99.8 \mathrm{kPa}\), calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is \(455 \mathrm{~mL}\) and its temperature is \(37^{\circ} \mathrm{C},\) calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose ( \(\left.\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2} ?\) (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). See Section 3.2 and Problem \(10.57 .\) )

A quantity of \(\mathrm{N}_{2}\) gas originally held at \(531.96 \mathrm{kPa}\) pressure in a 1.00 - \(\mathrm{L}\) container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5-\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(531.96 \mathrm{kPa}\) and \(26^{\circ} \mathrm{C}\) in a \(5.00-\mathrm{L}\) container is transferred to this same container. What is the total pressure in the new container?

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