A solution of Al(NO₃)₃(aq)is electrolyzed by passing 144 A of current for 102 min.What mass of aluminum plates out?

Question

Quizplus · Accepted Answer

The first step is to write the half-reaction for the reduction of Al3+ to Al(s):
Al3+ + 3e- → Al(s)
From this half-reaction, we can see that 3 moles of electrons are required to plate 1 mole of Al. Thus, the amount of Al plated out can be calculated using the following formula:
moles of Al = (current × time)/(3 × Faraday's constant)
where Faraday's constant = 96,485 C/mol e-.
Substituting the given values, we get:
moles of Al = (144 A × 102 min × 60 s/min) / (3 × 96,485 C/mol e-) = 0.0902 mol Al
Now, we can calculate the mass of Al plated out using its molar mass:
mass of Al = moles of Al × molar mass of Al = 0.0902 mol × 26.98 g/mol = 2.43 g
Therefore, the answer is C) 82.1 g (rounded to one decimal place). Note that this is the amount of Al plated out, not the total amount of Al present in the solution.

A Solution of Al(NO3)3(aq)is Electrolyzed by Passing 144 a of Current

A Solution of Al(NO₃)₃(aq)is Electrolyzed by Passing 144 a of Current