Copper (II) sulfate pentahydrate may be heated to drive off the water of hydration. If 5 g of water are produced, what was the original weight of the copper (II) sulfate pentahydrate (CuSO4 • 5H2O)? CuSO4 • 5H2O → CuSO4 + H2O. The answer given was 13.9g. Can someone explain how this answer is found?

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Eduard22sly Eduard22sly · Answer 1 · 2020-07-08T02:18:11+02:00

Answer:

13.9g of CuSO4•5H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CuSO4•5H2O → CuSO4 + 5H2O

Next, we shall determine the mass of CuSO4•5H2O heated and the mass of H2O produced from the balanced equation.

This is illustrated below

Molar mass of CuSO4• 5H2O = 63.5 + 32 + (16x4) + 5(2x1 + 16)

= 63.5 + 32 + 64 + 5(18) = 249.5g/mol

Mass of CuSO4•5H2O from the balanced equation = 1 x 249.5 = 249.5g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 5 x 18 = 90g

From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

Now, we can determine the mass of CuSO4•5H2O needed to produce 5g of H2O. This can be achieved as shown below:

From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

Therefore, Xg of CuSO4•5H2O will produce 5g of H2O i.e

Xg of CuSO4•5H2O = (249.5 x 5)/90

Xg of CuSO4•5H2O = 13.9g

Therefore, 13.9g of CuSO4•5H2O is needed to produce 5g of H2O.