%H2O = (12*molar mass H2O/molar mass KAl(SO)4)2.12H2O)*100 = ??
% anhydrous alum = 100 - %H2O
fraction anydrous alum = % anhydrous alum/100.
grams anhydrous alum in 100 g alum = 100g x fraction anhydrous alum.
By the way, there is no water, theoretical or otherwise, in anhydrous alum.
what will be the mass of angydrous Alum left over if one starts with 100.00 g of hydrated Alum?
please can you explain what fraction do I need to use. Based on the given information I can find the %theoretical of water in angydrous Alum. But I still don't understand how to solve for left over.
1 answer
1 answer