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If the starting weight of aluminum is 1.2g, what is the theoretical yield of alum in grams?
Also, what is the minimum volume of 1.5 M KOH that will be required to dissolve this weight?
Thanks for the help, exam soon and I am completely clueless on how to work this out.
1 Answer
- Roger the MoleLv 79 years agoFavorite Answer
Supposing your alum to be KAl(SO4)2·12H2O:
(1.2 g Al) / (26.98154 g Al/mol) x (1 mol alum/1 mol Al) x
(474.3902 g KAl(SO4)2·12H2O/mol) = 21 g KAl(SO4)2·12H2O
2 Al + 6 KOH → 3 H2 + 2 K3AlO3
(1.2 g Al) / (26.98154 g Al/mol) x (6/2) / (1.5 mol KOH/L) =
0.089 L KOH