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Chapter 10: Problem 60

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掳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 脳 Molar mass of C) = 40.08 + (2 脳 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掳C. Using the given information, the vapor pressure of water at 23掳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掳C = 296K) given. R is the ideal gas constant, which is 8.314 J/(mol路K) in SI units. However, we need the value of R in L路kPa/(mol路K) units so that we can find the volume in liters. 1 L路kPa = 100 J, so R = 8.314 J/(mol路K) 脳 (1 L路kPa / 100 J) = 0.0821 L路kPa/(mol路K). Now, let's calculate the volume of C2H2. V = (n 脳 R 脳 T) / P = (0.0238 mol 脳 0.0821 L路kPa/(mol路K) 脳 296 K) / 97.6 kPa = 0.595 L Thus, the volume of C2H2 that is collected over water at 23掳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鈥檚 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掳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鈥檚 Law and solve gas law problems efficiently.

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Most popular questions from this chapter

(a) The compound 1 -iodododecane is a nonvolatile liquid with a density of \(1.20 \mathrm{~g} / \mathrm{mL}\). The density of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\). What do you predict for the height of a barometer column based on 1 -iodododecane, when the atmospheric pressure is 749 torr? \((\mathbf{b})\) What is the pressure, in atmospheres, on the body of a diver if he is \(21 \mathrm{ft}\) below the surface of the water when the atmospheric pressure is 742 torr?

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathbf{a})\) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at 99.7 \(\mathrm{kPa}\) and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

(a) Are you more likely to see the density of a gas reported in \(\mathrm{g} / \mathrm{mL}, \mathrm{g} / \mathrm{L},\) or \(\mathrm{kg} / \mathrm{cm}^{3} ?(\mathbf{b})\) Which units are appropriate for expressing atmospheric pressures, \(\mathrm{N}, \mathrm{Pa}, \mathrm{atm}, \mathrm{kg} / \mathrm{m}^{2} ?\) (c) Which is most likely to be a gas at room temperature and ordinary atmospheric pressure, \(\mathrm{F}_{2}, \mathrm{Br}_{2}, \mathrm{~K}_{2} \mathrm{O} .\)

A mixture containing \(0.50 \mathrm{~mol} \mathrm{H}_{2}(g), 1.00 \mathrm{~mol} \mathrm{O}_{2}(g)\), and 3.50 \(\mathrm{mol} \mathrm{N}_{2}(g)\) is confined in a 25.0-L vessel at \(25^{\circ} \mathrm{C}\). (a) Calculate the total pressure of the mixture. (b) Calculate the partial pressure of each of the gases in the mixture.

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(3.03 \times 10^{8} \mathrm{~m}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL} ?(\mathbf{b})\) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) Calculate the total enthalpy change for complete combustion of the \(3.03 \times 10^{8} \mathrm{~m}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL} ;\) the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?
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