Answer:
3.36 grams Al°(s)
Explanation:
Given AlCl₃(s), determine the mass (grams) of Al°(s) produced from electrolysis of Aluminum Chloride at 10.0 amps for 1.00 hour.
AlCl₃(s) + 378.3°F (=192.4°C) => Al⁺³(l) + 3Cl⁻(l)
formula wt. Al° = 27g/mol
Faraday Constant (F°) = 96,500 amp·sec
? grams Al°(s) = 10.0amps x (1 mole e⁻/96,500amp-sec) x (1 mole Al°(s)/3 mole e⁻) x (27g Al°(s)/1 mole Al°(s)) x 3,600 sec = 3.36 grams Al°(s)
