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A 0.755-g sample of hydrated copper(II) sulfate, , was heated carefully until it had changed completely to anhydrous copper(II) sulfate (CuSO4) with a mass of 0.483 g. Determine the value of x. [This number is called the number of waters of hydration of copper(II) sulfate. It specifies the number of water molecules per formula unit of CuSO4 in the hydrated crystal.]

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Christian Garcia
Christian GarciaLv10
16 Nov 2020

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Related questions

When copper(II) sulfate pentahydrate (CuSO4

·

5 H2O) is heated, it decomposes to the dehydrated form. The waters of hydration are released from the solid crystal and form water vapor.

The hydrated form is medium blue, and the dehydrated solid is light blue.



The balanced equation is:

CuSO4 * 5 H2O(s) --> CuSO4(s) + 5 H2O(g)


If 4.47 g CuSO4 * 5 H2O is decomposed to CuSO4, predict the mass of the remaining light blue solid.

1.A hydrated form of copper sulfate (CuSO4·x H2O) is heated to drive off all the water. If we start with 9.19 g of hydrated salt and have 5.87 g of anhydrous CuSO4 after heating, find the number of water molecules associated with each CuSO4 formula unit.

2.Iron(II) Sulfate forms a blue-green hydrate with the formula FeSO4 ·n H2O(s). If this hydrate is heated to a high enough temperature, H2O(g) can be driven off, leaving the dirty yellow anhydrous salt FeSO4(s). A 18.300-g sample of the hydrate was heated to 300 °C. The resulting FeSO4(s) had a mass of 9.9993 g. Calculate the value of n in FeSO4 ·n H2O(s).

3.The combustion of ethane (C2H6) produces carbon dioxide and steam:

2C2H6 (G)+7O2(G)------->4CO2 (G)+6H2O (G)

How many moles of CO2 are produced when 5.15 mol of ethane are burned in an excess of oxygen?

--------MOLS?

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