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4 years ago

What is the best method for removing water from a hydrated compound?

Chemistry

2 answers:

Readme [11.4K]4 years ago

7 0

Answer:

Heating to remove water of crystallization
Passing through a drying agent (deliquescent or hygroscopic substances)

Explanation:

If solid/crystals, compound can be dried by applying heat to remove the water of crystallization.

Drying agents are also utile in removing moisture from compounds in a dessicator. Substances which absorb water from air can be used as drying agents for gases. For instance, $CaCl_{2}$ , CaO or silica gel is suitable for drying gases, depending on reactivity of the substance.

Hatshy [7]4 years ago

6 0

Answer:

Heating is the best method

Explanation:

When a mass of crystals containing Water of crystallization is heated sufiiciently to a high temperature, water vapor may be driven off the crystals by the action of heat alone.

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A chemist used 6.5 moles of water in this reaction. How many grams of water were used?

denpristay [2]

Answer:

$\boxed {\boxed {\sf 120 \ or \ 117 \ grams \ H_2O \ depending \ on \ significant \ figures }}$

Explanation:

We want to convert from moles of water to grams of water.

First, find the molar mass of water (H₂O) Look on the Periodic Table for the masses of hydrogen and oxygen.

Hydrogen (H): 1.008 g/mol
Oxygen (O): 15.999 g/mol

Next, add up the number of each element in water. The subscript of 2 comes after the H, so there are 2 moles of hydrogen.

2 Hydrogen: (1.008 g/mol*2) = 2.016 g/mol

Finally, add the molar mass of 2 hydrogen and 1 oxygen.

2.016 g/mol (2 Hydrogen) + 15.999 g/mol (1 oxygen)= 18.015 g/mol

Next, find the grams in 6.5 moles.

Use the molar mass we just found as a ratio.

$molar \ mass \ ratio: \frac{18.015 \ g \ H_2O}{1 \ mol \ H_2O}$

We want to find the grams in 6.5 moles. We can multiply the ratio above by 6.5

$6.5 \ mol \ H_2O * \frac{18.015 \ g \ H_2O}{1 \ mol \ H_2O}$

Multiply. Note that the moles of H₂O will cancel each other out.

$6.5 * \frac{18.015 \ g \ H_2O}{1}$

$6.5 * {18.015 \ g \ H_2O}$

$117.0975 \ g \ H_2O$

If we want to round to the technically correct significant figures, it would be 2 sig figs. The original measurement, 6.5, has 2 (6 and 5).

$\approx 120 \ g \ H_2O$

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3 years ago

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Roman55 [17]

Answer: 89.57 %

<em>Hope it helps</em>

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