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2. The same ratios can be used to predict the mass of the products given the mass of the reactants.a. A sample of copper (II) sulfate pentahydrate weighs 5.9 grams.How many grams of water would be released if all of the water is removed from the pentahydrate?Pls see pictures for details

2 The Same Ratios Can Be Used To Predict The Mass Of The Products Given The Mass Of The Reactantsa A Sample Of Copper II Sulfate Pentahydrate Weighs 59 GramsHow class=

Sagot :

Answer

2.12872 grams of water

Explanation

Given:

Mass of copper (II) sulfate pentahydrate sample = 5.9 grams

What to find:

The grams of water that would be released if all of the water is removed from the 5.9 grams sample of copper (II) sulfate pentahydrate.

Step-by-step solution:

The chemical formula of copper (II) sulfate pentahydrate is: CuSO₄.5H₂O

Molar mass of CuSO₄.5H₂O = 249.68 g/mol

Mass of water in 1 mole of CuSO₄.5H₂O = 5 x molar mass of water = 5 x 18.01528 g/mol = 90.0764 g/mol

the percentage by mass of water in 1 mole of CuSO₄.5H₂O is

[tex]\begin{gathered} \%\text{ }by\text{ }mass\text{ }of\text{ }water=\frac{Mass\text{ }of\text{ }water}{Molar\text{ }mass\text{ }of\text{ }CuSO₄.5H₂O}\times100\% \\ \\ \%\text{ }by\text{ }mass\text{ }of\text{ }water=\frac{90.0764}{249.68}\times100\% \\ \\ \%\text{ }by\text{ }mass\text{ }of\text{ }water=36.08\% \end{gathered}[/tex]

Hence, you can now use the % by mass of water in 1 mole of copper (II) sulfate pentahydrate to determine the mass of water that would be released in 5.9 grams of copper (II) sulfate pentahydrate as shown below.

[tex]\begin{gathered} Mass\text{ }of\text{ }water=\frac{36.08}{100}\times5.9\text{ }grams \\ \\ Mass\text{ }of\text{ }water=0.3608\times5.9grams \\ \\ Mass\text{ }of\text{ }water=2.12872\text{ }grams \end{gathered}[/tex]

Therefore, the grams of water that would be released if all of the water is removed from the 5.9 grams of copper (II) sulfate pentahydrate is 2.12872 grams

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