Question

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

Answer 1

2.12872 grams of water

Step-by-step 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

`\begin{gathered} \%\text{ }by\text{ }mass\text{ }of\text{ }water=\frac{Mass\text{ }of\text{ }water}{Molar\text{ }mass\text{ }of\text{ }CuSO₄.5H₂O}*100\% \\ \\ \%\text{ }by\text{ }mass\text{ }of\text{ }water=(90.0764)/(249.68)*100\% \\ \\ \%\text{ }by\text{ }mass\text{ }of\text{ }water=36.08\% \end{gathered}`

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.

`\begin{gathered} Mass\text{ }of\text{ }water=(36.08)/(100)*5.9\text{ }grams \\ \\ Mass\text{ }of\text{ }water=0.3608*5.9grams \\ \\ Mass\text{ }of\text{ }water=2.12872\text{ }grams \end{gathered}`

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