<span>Ans :
The chemical formula for aluminum sulfate is Alâ‚‚(SOâ‚„)â‚
6.00 mol Alâ‚‚(SOâ‚„)â‚
Al: 6.00 mol Alâ‚‚(SOâ‚„)â‚ x (2 mol Al / 1 mol Alâ‚‚(SOâ‚„)â‚ ) = 12.00 mol Al
S: 6.00 mol Alâ‚‚(SOâ‚„)â‚ x (3 mol S / 1 mol Alâ‚‚(SOâ‚„)â‚ ) = 18.00 mol S
O: 6.00 mol Alâ‚‚(SOâ‚„)â‚ x (12 mol O / 1 mol Alâ‚‚(SOâ‚„)â‚ ) = 72.00 mol O</span>
<span>This is rather a case of purification of impure copper or extraction of Cu from its alloy. You need to place the impure copper rod on the positive electrode (Anode) usualyy made of carbon rod, whereby oxidation reaction takes place: Cu (s) -------> Cu2+ (aq) + 2e- That is to say the impure cooper dissolves into solution. The copper (II) ions move to the negative electrode (cathode) usually made of pure copper rod. At the cathode, the Cu2+ ions are reduced : Cu2 (aq)+ + 2e- -----> Cu(s). That is to say the copper (II) ions are deposited as solid copper atoms onto the cu-rod electrode. In this way impure copper is deposited as pure copper onto the copper cathode</span>
The percent by mass of hydrogen in the compound is found by doing: (12.3/210.3) x 100 = 6%
Hope this will help
