Wednesday, February 8, 2012

Why wouldn't heating a hydrate in a crucible to calculate percentage of water be appropriate for all hydrates?

Because some of them, like CuCl2*2H2O, might be converted to cupric oxide.



CuCl2*2H2O(s) + heat ===%26gt; CuO(s) + 2HCl(g) + H2O(g)Why wouldn't heating a hydrate in a crucible to calculate percentage of water be appropriate for all hydrates?
Some hydrates would completely decompose instead of becoming anhydrous.

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