Question #a1a19

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Question #a1a19

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Answer 1 · 2017-10-03T01:17:51

Depending on what your teacher is looking for, you might be overthinking this one.

Explanation:

Immediate after heating the copper sulfate forms a dry, light blue solid.

Right after the heating is complete the reaction is:

$C u S O_{4} \cdot 5 H_{2} O (s) \to C u S O_{4} (s) + 5 H_{2} O (g)$ $CuSO_4*5H_2O(s) -> CuSO_4(s) + 5H_2O(g)$

The addition of water reverses the removal of water.

$C u S O_{4} (s) + 5 H_{2} O (l) \to C u S O_{4} \cdot 5 H_{2} O (s)$ $CuSO_4(s) + 5H_2O(l) -> CuSO_4*5H_2O(s)$

If you add exactly 5 parts water for every one part copper sulfate the solid will return to being $C u S O_{4} \cdot 5 H_{2} O (s)$ $CuSO_4*5H_2O(s)$ .

Adding any extra water will cause the formation of a copper sulfate solution of $C u S O_{4} \cdot 5 H_{2} O (a q)$ $CuSO_4*5H_2O(aq)$ .

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