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kvv77 [185]
1 year ago
12

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

eleased?
Chemistry
2 answers:
Ivan1 year ago
7 0

Mass of water released =

92.8 g CuSO_{4}.5H_{2}O×\frac{5 * 18 g H_{2}O}{249.68 g CuSO_{4}.5H_{2}O}

= 33.45 g H_{2}O

babymother [125]1 year ago
5 0

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

<u>Explanation:</u>

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

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

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

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

So, in 92.8 grams of the compound, \frac{5\times 18}{249.55}\times 92.8=33.46g of water molecule.

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

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The estimated feed rate of logs is 14.3 logs/min.

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