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olchik [2.2K]
3 years ago
8

What mass of anhydrous copper (II) sulfate can be obtained by the heating of 125.00 grams of copper (II) sulfate pentahydrate?

Chemistry
1 answer:
RSB [31]3 years ago
4 0

Answer:

79.91 g

Explanation:

First we <u>calculate the number of moles of 125.00 grams of copper (II) sulfate pentahydrate </u>(CuSO₄·5H₂O), using<em> its molar mass</em>:

  • Molar Mass of CuSO₄·5H₂O = (Molar Mass of CuSO₄) + 5*(Molar Mass of H₂O)
  • Molar Mass of CuSO₄·5H₂O = 249.68 g/mol
  • moles CuSO₄·5H₂O = 125.00 g ÷ 249.68 g/mol = 0.501 mol CuSO₄·5H₂O

In the reaction, CuSO₄·5H₂O turns into CuSO₄.

So now <u>we convert 0.501 moles of CuSO₄ (anhydrous copper (II) sulfate) into grams</u>, using the <em>molar mass of CuSO₄</em>:

  • 0.501 mol CuSO₄ * 159.609 g/mol = 79.91 g CuSO₄
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