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Chapter 18: Problem 130

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 \(257 \mathrm{~L}\) \(\mathrm{Cl}_{2}(\mathrm{~g})\), with both volumes measured at \(50 .{ }^{\circ} \mathrm{C}\) and \(2.50 \mathrm{~atm} ?\)

Short Answer

Expert verified
In the electrolysis of a sodium chloride solution, 257 L of H₂ (g) is produced in the same time it takes to produce 257 L of Cl₂(g), with both volumes measured at 50.°C and 2.50 atm.

Step by step solution

01

Write the balanced chemical equation for the electrolysis of sodium chloride solution.

The balanced equation for the electrolysis of sodium chloride solution is as follows: \[2 \mathrm{NaCl} (aq) + 2 \mathrm{H}_{2}O (l) \rightarrow 2 \mathrm{NaOH} (aq) + \mathrm{Cl}_{2} (g) + \mathrm{H}_{2} (g)\] The stoichiometry from the balanced equation shows that one mole of Cl₂ gas is produced for every one mole of H₂ gas.
02

Find the amount of moles of Cl₂ gas produced.

We will use the Ideal Gas Law formula \(PV = nRT\) to find the moles of Cl₂ gas produced, where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is temperature in Kelvin. Given: Volume of Cl₂ = 257 L Pressure = 2.50 atm Temperature = 50 °C = 323 K (To convert from Celsius to Kelvin, add 273) Universal Gas Constant R = 0.0821 L atm / (mol K) Plugging these values into the Ideal Gas Law formula: \(2.50 \cdot 257 = n \cdot 0.0821 \cdot 323\) Now, we will solve the equation for the number of moles (n) of Cl₂.
03

Calculate the moles of Cl₂ produced.

Solving for the number of moles (n) of Cl₂ produced: \(n_{Cl2} = \frac{2.50 \cdot 257}{0.0821 \cdot 323} = 25.92~\text{moles}\) We find that 25.92 moles of Cl₂ gas are produced during the electrolysis of the sodium chloride solution.
04

Find the moles of H₂ gas produced.

In the balanced equation, for every mole of Cl₂ produced, one mole of H₂ is also produced: \(n_{H2} = n_{Cl2}\) \(n_{H2} = 25.92~\text{moles}\) So, 25.92 moles of H₂ gas are produced.
05

Calculate the volume of H₂ gas produced.

Now that we have the moles of H₂ (n), we can use the Ideal Gas Law formula to find the volume of H₂ gas produced under the given conditions: \(PV = nRT\) \(V_{H2} = \frac{n_{H2} \cdot R \cdot T} {P}\) Plugging the values into the formula: \(V_{H2} = \frac {25.92 \cdot 0.0821 \cdot 323}{2.50} = 257~\text{L}\) Thus, the volume of H₂ gas produced under these conditions is 257 L.

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Most popular questions from this chapter

The following standard reduction potentials have been determined for the aqueous chemistry of indium: $$\begin{array}{ll}\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V}\end{array}$$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ b. What is \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q)\) ?

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