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Chapter 6: Problem 26

A 34.0 L cylinder contains \(305 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\) at \(22^{\circ} \mathrm{C} .\) How many grams of \(\mathrm{O}_{2}(\mathrm{g})\) must be released to reduce the pressure in the cylinder to 1.15 atm if the temperature remains constant?

Short Answer

Expert verified
The cylinder needs to release 253.44 g of O2 gas to reduce the pressure to 1.15 atm.

Step by step solution

01

Calculate the Initial Number of Moles

First, calculate the initial number of moles of O2 gas using the Ideal Gas Law. The initial pressure is 1 atm by virtue of it being a standard container, and the volume is 34.0 L. The temperature is 22 degrees Celsius, which converted to Kelvin is 295.15 K, and the gas constant R is 0.0821 L∙atm/K∙mol. Thus, \(n_1 = PV / RT = (1 atm × 34.0 L) / (0.0821 L∙atm/K∙mol × 295.15 K) = 1.402 mol\) . We also know that the O2 cylinder initially has 305 g of O2, which corresponds to \(305 g / 32.00 g/mol = 9.53 mol\).
02

On Determine the Final Number of Moles

Next the number of moles of O2 gas that would create a pressure of 1.15 atm is calculated. With the volume (34.0 L), temperature (295.15 K), and gas constant (0.0821 L∙atm/K∙mol) unchanged, the new pressure is 1.15 atm. Hence, \(n_2 = PV / RT = (1.15 atm × 34.0 L) / (0.0821 L∙atm/K∙mol × 295.15 K) = 1.611 mol\).
03

Calculate the Released O2

The initial number of moles minus the final number of moles gives the number of moles of O2 gas that must be released to reduce the pressure to 1.15 atm. Therefore, \(n = n_1 - n_2 = 9.53 mol - 1.61 mol = 7.92 mol\).
04

Convert Moles into Grams

The last step is to convert the number of moles (7.92 mol) into grams using the molar mass of O2 (32.00 g/mol). Therefore, the mass of O2 that needs to be released is \(7.92 mol × 32.00 g/mol = 253.44 g\).

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Most popular questions from this chapter

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) has a volume of \(26.7 \mathrm{L}\) at 762 Torr. What is the new volume if, with the temperature and amount of gas held constant, the pressure is (a) lowered to 385 Torr; (b) increased to 3.68 atm?

At elevated temperatures, solid sodium chlorate \(\left(\mathrm{NaClO}_{3}\right)\) decomposes to produce sodium chloride, \(\mathrm{NaCl},\) and \(\mathrm{O}_{2}\) gas. A \(0.8765 \mathrm{g}\) sample of impure sodium chlorate was heated until the production of oxygen ceased. The oxygen gas was collected over water and occupied a volume of \(57.2 \mathrm{mL}\) at \(23.0^{\circ} \mathrm{C}\) and 734 Torr. Calculate the mass percentage of \(\mathrm{NaClO}_{3}\) in the original sample. Assume that none of the impurities produce oxygen on heating. The vapor pressure of water is 21.07 Torr at \(23^{\circ} \mathrm{C}\).

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