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Chapter 5: Problem 68

A 2.5 -L flask at \(15^{\circ} \mathrm{C}\) contains a mixture of \(\mathrm{N}_{2}, \mathrm{He}\) and Ne at partial pressures of 0.32 atm for \(\mathrm{N}_{2}\), 0.15 atm for \(\mathrm{He},\) and 0.42 atm for \(\mathrm{Ne} .\) (a) Calculate the total pressure of the mixture. (b) Calculate the volume in liters at STP occupied by He and Ne if the \(\mathrm{N}_{2}\) is removed selectively.

Short Answer

Expert verified
The total pressure of the mixture is 0.89 atm. The volume occupied by He and Ne at STP (if the N2 is selectively removed) is approximately 1.1 liters.

Step by step solution

01

Calculate the total pressure

The total pressure of a mixture of gases can be calculated simply by adding up the partial pressures of each gas in the mixture. Therefore, the total pressure \( P_t \) can be found using the equation: \(P_t = P_{N2} + P_{He} + P_{Ne}\). Substituting the given values and calculating, \(P_t = 0.32 atm + 0.15 atm + 0.42 atm = 0.89 atm \)
02

Calculate the Moles of He and Ne

Next, we calculate the number of moles of the gases He and Ne using the ideal gas law, rearranged to solve for n: \(n = P*V/R*T\). Here \(R = 0.0821 L.atm/mol.K\) which is a common value for the gas constant and the temperature T is converted to Kelvin (15 C = 288 K). So the moles of He and Ne can be calculated as: \(n_{He} = P_{He}*V/R*T \rightarrow n_{He} = 0.15*2.5/0.0821*288 = 0.0129 mol\), \(n_{Ne} = P_{Ne}*V/R*T \rightarrow n_{Ne} = 0.42*2.5/0.0821*288 = 0.0361 mol\)
03

Calculate the volume

After calculating the moles of He and Ne, we next calculate the volume they would occupy at STP. We know that any number of moles of gas at STP occupies 22.4 L. Hence the volume of He and Ne at STP can be calculated as: \(V_{He} = n_{He}*22.4 = 0.0129*22.4 = 0.2896 L\), \(V_{Ne} = n_{Ne}*22.4 = 0.0361*22.4 = 0.8086 L\). Hence the total volume \(V_t = V_{He} + V_{Ne} = 0.2896 + 0.8086 = 1.0982 L \)

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Most popular questions from this chapter

Commercially, compressed oxygen is sold in metal cylinders. If a 120-L cylinder is filled with oxygen to a pressure of 132 atm at \(22^{\circ} \mathrm{C},\) what is the mass (in grams) of \(\mathrm{O}_{2}\) present? How many liters of \(\mathrm{O}_{2}\) gas at 1.00 atm and \(22^{\circ} \mathrm{C}\) could the cylinder produce? (Assume ideal behavior.)

In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g) $$ If \(5.97 \mathrm{~g}\) of glucose are reacted and \(1.44 \mathrm{~L}\) of \(\mathrm{CO}_{2}\) gas are collected at \(293 \mathrm{~K}\) and \(0.984 \mathrm{~atm},\) what is the percent yield of the reaction?

A sample of zinc metal reacts completely with an excess of hydrochloric acid: $$ \operatorname{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ The hydrogen gas produced is collected over water at \(25.0^{\circ} \mathrm{C}\) using an arrangement similar to that shown in Figure \(5.15 .\) The volume of the gas is \(7.80 \mathrm{~L},\) and the pressure is \(0.980 \mathrm{~atm} .\) Calculate the amount of zinc metal in grams consumed in the reaction. (Vapor pressure of water at \(25^{\circ} \mathrm{C}=23.8 \mathrm{mmHg} .\) )

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