The energy change ∆H associated with the reaction NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g) is +81 kJ. NBr3 is arranged with all o
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Answer:
There are present 5,5668 moles of water per mole of CuSO₄.
Explanation:
The mass of CuSO₄ anhydrous is:
23,403g - 22,652g = 0,751g.
mass of crucible+lid+CuSO₄ - mass of crucible+lid
As molar mass of CuSO₄ is 159,609g/mol. The moles are:
0,751g × = 4,7052x10⁻³ moles CuSO₄
Now, the mass of water present in the initial sample is:
23,875g - 0,751g - 22,652g = 0,472g.
mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid
As molar mass of H₂O is 18,02g/mol. The moles are:
0,472g × = 2,6193x10⁻² moles H₂O
The ratio of moles H₂O:CuSO₄ is:
2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668
That means that you have <em>5,5668 moles of water per mole of CuSO₄.</em>
I hope it helps!