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Chapter 1: Problem 24

Molten \(\mathrm{AlCl}_{3}\) is electrolyzed with a constant current of 5.00 amperes over a period of 600.0 seconds. Which of the following expressions is equal to the maximum mass of Al(s) that plates out? (1 faraday = 96,500 coulombs) (A) \(\frac{(600)(5.00)}{(96,500)(3)(27.0)}\) grams (B) \(\frac{(600)(5.00)(3)(27.0)}{(96,500)}\) grams (C) \(\frac{(600)(5.00)(27.0)}{(96,500)(3)}\) grams (D) \(\frac{(96,500)(3)(27.0)}{(600)(5.00)}\) grams

Short Answer

Expert verified
The maximum mass of Al that plates out is represented by the expression in option (A).

Step by step solution

01

Stoichiometry of the reaction

In the production of metallic Al from AlCl3, the reaction is: \[Al^{3+} + 3e^- \rightarrow Al(s)\] This indicates that three moles of electrons (or 3 faradays) are needed to produce one mole of Al.
02

Calculate total charge

The total charge (Q) passed in the electrolysis can be calculated from the provided current (I) and the time (t). Use the formula \[Q = It\] where I is given as 5.00 Amperes and t is 600.0 seconds. This gives \[Q = (5.00 A)(600.0 s)\] coulombs.
03

Apply Faraday's Law

Applying Faraday’s Law, for the deposition of one mole of Al, three faradays or 3*96,500 coulombs are needed. Hence, calculate the mass using the formula \[mass = \frac{Q}{F}*\frac{1}{n}*M\] where F is Faraday’s constant (96,500 C), n is the number of electrons involved (3 for aluminium), M is the molar mass of Al (27.0 g/mol). Substituting the values the mass becomes: \[mass = \frac{(5.00 A)(600.0 s)}{(96,500 C/mol)(3 mol)(27.0 g/mol)}\]
04

Compare with given options

On close inspection, the correct expression matches with option A. Therefore, the maximum mass of Al that plates out is given by the expression in option A.

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Most popular questions from this chapter

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The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) What is the expression for the equilibrium constant, \(K_{\mathrm{c}} ?\) (A) \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}\) (B) \(\frac{2\left[\mathrm{SO}_{3}\right]}{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}\) (C) \(\frac{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}{\left[\mathrm{SO}_{3}\right]^{2}}\) (D) \(\frac{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}{2\left[\mathrm{SO}_{3}\right]}\)

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . If some of the solution evaporates overnight, which of the following will occur? (A) The mass of the solid and the concentration of the ions will stay the same. (B) The mass of the solid and the concentration of the ions will increase. (C) The mass of the solid will decrease, and the concentration of the ions will stay the same. (D) The mass of the solid will increase, and the concentration of the ions will stay the same.

\(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)\) Based on the information given in the table below, what is \(\Delta H^{\circ}\) for the above reaction? \(\begin{array}{cc}{\text { Bond }} & {\text { Average bond energy }(\mathrm{kJ} / \mathrm{mol})} \\ {\mathrm{H}-\mathrm{H}} & {500} \\\ {\mathrm{O}=\mathrm{O}} & {500} \\ {\mathrm{O}-\mathrm{H}} & {500}\end{array}\) (A) \(-2,000 \mathrm{kJ}\) (B) \(-500 \mathrm{kJ}\) (C) \(+1,000 \mathrm{kJ}\) (D) \(+2,000 \mathrm{kJ}\)
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