Question

A 92.8 gram sample of CuSO4•5H2O.(copper sulfate pentahydrate) is heated until water is released. How many grams of water were released?

Answer 1

Mass of water released =

`92.8 g CuSO_(4).5H_(2)O`×
`(5 * 18 g H_(2)O)/(249.68 g CuSO_(4).5H_(2)O)`

= 33.45 g
`H_(2)O`

Answer 2

Answer: The mass of water released from the given amount of copper sulfate is 33.46 grams.

Step-by-step explanation:

We are given a compound having chemical formula
`CuSO_4.5H_2O`

The molar mass of this compound =
`[63.55+32+(4* 16)+5(16+(2* 1))]=249.55g/mol`

Mass of water molecule =
`[16+(2* 1)]=18g/mol`

In 249.55 grams of the compound,
`(5* 18)g` of water molecule is present.

So, in 92.8 grams of the compound,
`(5* 18)/(249.55)* 92.8=33.46g` of water molecule.

Hence, the mass of water released from the given amount of copper sulfate is 33.46 grams.