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

2 Answers

6 votes

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

User DarVar
by
8.4k points
4 votes

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.

User Nitin Labhishetty
by
7.8k points
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