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Chapter 17: Problem 132

In the electrolysis of a sodium chloride solution, what volume of \(\mathrm{H}_{2}(g)\) is produced in the same time it takes to produce \(257 \mathrm{L}\) \(\mathrm{Cl}_{2}(g),\) with both volumes measured at \(50 .^{\circ} \mathrm{C}\) and 2.50 atm?

Short Answer

Expert verified
The volume of hydrogen gas (\(H_2\)) produced in the same time it takes to produce 257 L of chlorine gas (\(Cl_2\)), with both volumes measured at 50°C and 2.50 atm, is 257 L.

Step by step solution

01

Write the balanced chemical equation for the electrolysis process

The electrolysis of sodium chloride solution results in the production of hydrogen gas, chlorine gas, and sodium hydroxide. The balanced chemical equation for this process is: 2NaCl(aq) + 2H₂O(l) -> 2NaOH(aq) + H₂(g) + Cl₂(g)
02

List the given information and find the stoichiometric ratio of hydrogen and chlorine gases

We know the volume of chlorine gas produced is 257 L, and the conditions are 50°C and 2.50 atm. We also note from the balanced chemical equation that the stoichiometric ratio of hydrogen gas to chlorine gas is 1:1, as indicated by the coefficients.
03

Convert temperature to Kelvin

To use the ideal gas law, we must convert the temperature from Celsius to Kelvin: Temperature in Kelvin = Temperature in Celsius + 273.15 T(K) = 50°C + 273.15 = 323.15 K
04

Use stoichiometric ratio to find the volume of hydrogen gas produced

Since the ratio of hydrogen gas to chlorine gas is 1:1, the volume of hydrogen gas produced is the same as the volume of chlorine gas produced: Volume of hydrogen gas = Volume of chlorine gas = 257 L
05

Use the ideal gas law to determine the number of moles for chlorine gas

We can use the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature, to find the number of moles for chlorine gas: n(Cl₂) = PV / RT = (2.50 atm)(257 L) / (0.0821 L*atm/mol*K)(323.15 K) = 24.39 mol
06

Use the stoichiometric ratio to find the number of moles of hydrogen gas

Now that we know the number of moles of chlorine gas, we can use the stoichiometric ratio to determine the number of moles of hydrogen gas, since the ratio is 1:1: n(H₂) = n(Cl₂) = 24.39 mol
07

Use the ideal gas law to determine the volume of hydrogen gas

Finally, we use the ideal gas law again to find the volume of hydrogen gas, now that we have the number of moles: V(H₂) = n(H₂)RT / P = (24.39 mol)(0.0821 L*atm/mol*K)(323.15 K) / (2.50 atm) = 257 L Therefore, the volume of hydrogen gas produced in the same time it takes to produce 257 L of chlorine gas, with both volumes measured at 50°C and 2.50 atm, is 257 L.

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Most popular questions from this chapter

It took 2.30 min using a current of 2.00 A to plate out all the silver from 0.250 L of a solution containing Ag \(^{+} .\) What was the original concentration of \(\mathrm{Ag}^{+}\) in the solution?

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Which of the following statement(s) is/are true? a. Copper metal can be oxidized by \(\mathrm{Ag}^{+}\) (at standard conditions). b. In a galvanic cell the oxidizing agent in the cell reaction is present at the anode. c. In a cell using the half reactions \(A l^{3+}+3 e^{-} \longrightarrow A l\) and \(\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg},\) aluminum functions as the anode. d. In a concentration cell electrons always flow from the compartment with the lower ion concentration to the compartment with the higher ion concentration. e. In a galvanic cell the negative ions in the salt bridge flow in the same direction as the electrons.

Consider a concentration cell that has both electrodes made of some metal M. Solution \(A\) in one compartment of the cell contains \(1.0 \mathrm{M} \mathrm{M}^{2+}\). Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathbf{M}^{2+}(a q)+\mathbf{S O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 V at 25 \(^{\circ}\) C. Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\)

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