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A hydrated form of copper sulfate (CuSO4Ā·x H2O) is heated to drive off all the water. If we start with 8.79 g of hydrated salt and have 6.57 g of anhydrous CuSO4 after heating, find the number of water molecules associated with each CuSO4 formula unit.

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Answer:

3 water molecules

Explanation:

The molar ratio between CuSOā‚„ and Hā‚‚O needs to be calculated.

The mass of water that was removed by heating is calculated, then converted to moles. The molecular weight of water is 18.02g/mol.

(8.97 g - 6.57 g) = 2.22 g Hā‚‚O

(2.22 g)/(18.02g/mol) = 0.1232 mol Hā‚‚O

The moles of anhydrous copper sulfate are calculated. The molecular weight is 159.609 g/mol

(6.57 g)/(159.609 g/mol) = 0.04116 mol CuSOā‚„

Now the molar ratio between CuSOā‚„ and Hā‚‚O can be calculated:

0.1232 mol Hā‚‚O Ć· 0.04116 mol CuSOā‚„ = 3 Hā‚‚O / CuSOā‚„

The molar ratios determine how many moles of product are created from a given amount of reactant, as well as how many moles of one reactant are required to thoroughly react with another reactant.

  • It is necessary to compute the molar ratio of [tex]CuSO_{4}[/tex] and [tex]H_{2} O[/tex].
  • The mass of water removed by heating is determined, and the moles are transformed. Water has a molecular weight of 18.02g/mol.

[tex](8.97 g - 6.57 g) = 2.22 g H_{2} O(2.22 g)/(18.02g/mol) = 0.1232 mol H_{2} O[/tex]

  • Calculate the moles of anhydrous copper sulfate. 159.609 g/mol is the molecular weight.

[tex](6.57 g)/(159.609 g/mol) = 0.04116 mol CuSO_{4}[/tex]

  • The molar ratio of [tex]CuSO_{4}[/tex] to [tex]H_{2} O[/tex] can now be calculated:

[tex]0.1232 mol H_{2} O / 0.04116 mol CuSO_{4} = 3 H_{2} O / CuSO_{4}[/tex]

Thus, the molar ratio is 3:1 meaning 3 water molecules are associated.

Learn more about molar ratios: https://brainly.com/question/14665653

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