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Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), can be prepared by the reaction of calcium carbide with water: $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g)$$ Calculate the volume of \(\mathrm{C}_{2} \mathrm{H}_{2}\) that is collected over water at \(23^{\circ} \mathrm{C}\) by reaction of \(1.524 \mathrm{~g}\) of \(\mathrm{CaC}_{2}\) if the total pressure of the gas is \(100.4 \mathrm{kPa}\). (The vapor pressure of water is tabulated in Appendix B.)

Short Answer

Expert verified
The volume of C2H2 that is collected over water at 23°C and 100.4 kPa is approximately 0.595 liters.

Step by step solution

01

Convert the mass of CaC2 to moles

Given, mass of CaC2 = 1.524 g. Calculate the molar mass of CaC2. Molar mass of CaC2 = Molar mass of Ca + (2 × Molar mass of C) = 40.08 + (2 × 12.01) = 64.10 g/mol Now, let's calculate the moles of CaC2. Moles of CaC2 = mass of CaC2 / molar mass of CaC2 = 1.524 g / 64.10 g/mol = 0.0238 mol
02

Use stoichiometry to find the number of moles of C2H2 produced

From the balanced chemical equation, we see that 1 mole of CaC2 reacts with water to produce 1 mole of C2H2. Therefore, the number of moles of C2H2 produced is equal to the number of moles of CaC2. Moles of C2H2 = 0.0238 mol
03

Use the vapor pressure of water to find the partial pressure of C2H2

The total pressure of the gas is given as 100.4 kPa, and we need to find the partial pressure of C2H2. First, we need to find the vapor pressure of water at 23°C. Using the given information, the vapor pressure of water at 23°C is 2.8 kPa. Now, using Dalton's law, we can find the partial pressure of C2H2. Partial pressure of C2H2 = Total pressure - Vapor pressure of water = 100.4 kPa - 2.8 kPa = 97.6 kPa
04

Use the ideal gas law to find the volume of C2H2

Now, let's use the ideal gas law (PV = nRT) to find the volume of C2H2. We have the partial pressure of C2H2, number of moles of C2H2, and the temperature (23°C = 296K) given. R is the ideal gas constant, which is 8.314 J/(mol·K) in SI units. However, we need the value of R in L·kPa/(mol·K) units so that we can find the volume in liters. 1 L·kPa = 100 J, so R = 8.314 J/(mol·K) × (1 L·kPa / 100 J) = 0.0821 L·kPa/(mol·K). Now, let's calculate the volume of C2H2. V = (n × R × T) / P = (0.0238 mol × 0.0821 L·kPa/(mol·K) × 296 K) / 97.6 kPa = 0.595 L Thus, the volume of C2H2 that is collected over water at 23°C and 100.4 kPa is approximately 0.595 liters.

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Key Concepts

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

Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships of reactants and products in a chemical reaction. It helps you measure how much of each substance you need or produce in a reaction.

To solve stoichiometry problems, follow these steps:
  • Write the balanced chemical equation. Ensure all atoms are balanced on both sides.
  • Use the molar masses of reactants and products to convert between grams and moles.
  • Apply the molar ratio from the balanced equation to find moles of desired substances.
In our exercise, we start by converting the mass of calcium carbide ( CaC_2 ) into moles using its molar mass. Then, using the chemical equation, determine the moles of acetylene ( C_2H_2 ) produced. This stepwise approach ensures accurate calculations.
Dalton's Law
Dalton’s Law of Partial Pressures states that in a mixture of gases, the total pressure is equal to the sum of the partial pressures of each individual gas. This is crucial when dealing with gases collected over water.

In our problem, the gas collected is a mixture of acetylene and water vapor. To find the pressure exerted by just the acetylene, subtract the water vapor pressure from the total pressure.
  • Total Pressure = Partial Pressure of gas + Vapor Pressure of water
  • Partial Pressure of C_2H_2 = Total Pressure - Vapor Pressure of water
By applying this principle, we can calculate the pressure exerted solely by acetylene gas, which is then used in further calculations.
Vapor Pressure
Vapor pressure is the pressure exerted by the vapor in equilibrium with its liquid at a given temperature. It's important in calculations involving gases collected over liquids because the liquid contributes to the total pressure with its vapor.

In our exercise, you need to know the vapor pressure of water at the reaction temperature (23°C), which is 2.8 kPa.
  • This is subtracted from the total pressure to find the pressure of the gas alone.
  • Vapor pressure changes with temperature, so always use the correct value for your specific conditions.
Understanding vapor pressure is key in accurately determining gas pressures in a reaction.
Partial Pressure
Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. Each gas contributes to the total pressure proportionally to its amount.

When calculating partial pressures, the total pressure is distributed among all gases present. For acetylene collected over water:
  • First, determine the pressure contributed by water vapor.
  • Subtract this from the total pressure to calculate acetylene's partial pressure.
Mastering the concept of partial pressure helps you accurately apply Dalton’s Law and solve gas law problems efficiently.

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Most popular questions from this chapter

Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at \(80^{\circ} \mathrm{C}\) if 1.00 mol occupies \(33.3 \mathrm{~L}\), assuming that (a) \(\mathrm{CCl}_{4}\) obeys the ideal-gas equation; (b) \(\mathrm{CCl}_{4}\) obeys the van der Waals equation. (Values for the van der Waals constants are given in Table \(10.3 .\) ) (c) Which would you expect to deviate more from ideal behavior under these conditions, \(\mathrm{Cl}_{2}\) or \(\mathrm{CCl}_{4} ?\) Explain.

In the United States, barometric pressures are generally reported in inches of mercury (in. Hg). On a beautiful summer day in Chicago, the barometric pressure is 30.45 in. \(\mathrm{Hg}\). (a) Convert this pressure to torr. (b) Convert this pressure to atm.

The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in our bodies produces \(\mathrm{CO}_{2}\), which is expelled from our lungs as a gas: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at normal body temperature, \(37^{\circ} \mathrm{C},\) and \(101.33 \mathrm{kPa}\) when \(10.0 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at \(100 \mathrm{kPa}\) and \(298 \mathrm{~K},\) to completely oxidize \(15.0 \mathrm{~g}\) of glucose.

Carbon dioxide, which is recognized as the major contributor to global warming as a "greenhouse gas," is formed when fossil fuels are combusted, as in electrical power plants fueled by coal, oil, or natural gas. One potential way to reduce the amount of \(\mathrm{CO}_{2}\) added to the atmosphere is to store it as a compressed gas in underground formations. Consider a 1000-megawatt coal-fired power plant that produces about \(6 \times 10^{6}\) tons of \(\mathrm{CO}_{2}\) per year. (a) Assuming ideal-gas behavior, \(101.3 \mathrm{kPa}\), and \(27^{\circ} \mathrm{C}\), calculate the volume of \(\mathrm{CO}_{2}\) produced by this power plant. (b) If the \(\mathrm{CO}_{2}\) is stored underground as a liquid at \(10^{\circ} \mathrm{C}\) and \(12.16 \mathrm{MPa}\) and a density of \(1.2 \mathrm{~g} / \mathrm{cm}^{3},\) what volume does it possess? (c) If it is stored underground as a gas at \(30^{\circ} \mathrm{C}\) and \(7.09 \mathrm{MPa}\), what volume does it occupy?

(a) Place the following gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{O}_{2}, \mathrm{Ar}, \mathrm{CO}, \mathrm{HCl}, \mathrm{CH}_{4} \cdot(\mathbf{b})\) Calculate the rms speed of CO molecules at \(25^{\circ} \mathrm{C} .(\mathbf{c})\) Calculate the most probable speed of an argon atom in the stratosphere, where the temperature is \(0^{\circ} \mathrm{C}\).

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