How do you write the balanced equation for the hydration of $C u S O_{4}$ $CuSO_4$ ?

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Answer 1 · 2016-11-15T23:21:33

${CuSO}_{4} (s) + {5H}_{2} O(l) \to {CuSO}_{4} \cdot {5H}_{2} O(s)$ $"CuSO"_4"(s)" + "5H"_2"O(l)" → "CuSO"_4·"5H"_2"O(s)"$

The formula for the hydrated ${CuSO}_{4}$ $"CuSO"_4$ is ${CuSO}_{4} \cdot {5H}_{2} O$ $"CuSO"_4·"5H"_2"O"$ .

The unbalanced equation is

${CuSO}_{4} (s) + H_{2} O(l) \to {CuSO}_{4} \cdot {5H}_{2} O(s)$ $"CuSO"_4"(s)" + "H"_2"O(l)" → "CuSO"_4·"5H"_2"O(s)"$

To balance it, put a 5 in front of $H_{2} O$ $"H"_2"O"$ .

This gives the balanced equation

${CuSO}_{4} (s) + {5H}_{2} O(l) \to {CuSO}_{4} \cdot {5H}_{2} O(s)$ $"CuSO"_4"(s)" + "5H"_2"O(l)" → "CuSO"_4·"5H"_2"O(s)"$

How do you write the balanced equation for the hydration of CuSO4CuSO_4?