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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°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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Most popular questions from this chapter

A rigid vessel containing a \(3: 1 \mathrm{~mol}\) ratio of carbon dioxide and water vapor is held at \(200^{\circ} \mathrm{C}\) where it has a total pressure of \(202.7 \mathrm{kPa}\). If the vessel is cooled to \(10^{\circ} \mathrm{C}\) so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.

The highest barometric pressure ever recorded was 823.7 torr at Agata in Siberia, Russia on December 31,1968 . Convert this pressure to (a) atm, (b) \(\mathrm{mm} \mathrm{Hg}\), (c) pascals, (d) bars, (e) psi.

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.

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