Approximately how many moles of al3+ are reduced when 0.1 faraday of charge passes through a cell during the production of al? (note: assume there is excess al3+ available and that al3+ is reduced to al metal only.)
a. 0.033 molb. 0.050 molc. 0.067 mold. 0.10 mol

0.033 mol

Explanation:

1 faraday is require to  1 mol of electron

0.1 faraday is required for 0.1 mol of electron

3 mol of electron is required for 1 mole Al3+ to Al

0.1 mole of electron will therefore need ( 0.1 mol × 1 mol ) / 3 mol = 0.033 mol

the answer would be a

lol for the pointsss