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3 years ago

The following data is given to find the formula of a Hydrate:

Chemistry

1 answer:

umka21 [38]3 years ago

3 0

The masses can be found by substractions:

Mass of CaSO₄.H2O (hydrate):

16.05 g - 13.56 g = 2.49 g

Mass of CaSO₄ anhydrate:

15.07 g - 13.56 g = 1.51 g

The mass of water is equal to the difference between the mass of the hydrate and the mass of the anhydrate:

2.49 g - 1.51 g = 0.98 g

The percent of water is found by the formula:

massWater ÷ massHydrate * 100%

0.98 g ÷ 2.49 g * 100% = 39.36%

The mole of water is calculated using water's molecular weight (18g/mol):

0.98 g ÷ 18 g/mol = 0.054 mol water

A similar procedure is made for the mole of salt (CaSO₄ = 136.14 g/mol)

1.51 g ÷ 136.14 g/mol = 0.011 mol CaSO₄

The ratio of mole of water to mole of anhydrate is:

0.054 mol water / 0.011 mol CaSO₄ = 0.49

In other words the molecular formula for the hydrate salt is CaSO₄·0.5H₂O

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The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound Amount

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Answer: The value of equilibrium constant for given equation is $1.0\times 10^{-4}$

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

For hydrogen gas:

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M$

For water:

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M$

For the given chemical equation:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression of equilibrium constant for above equation follows:

$K_{eq}=\frac{[H_2O]^3}{[H_2]^3}$

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}$

Hence, the value of equilibrium constant for given equation is $1.0\times 10^{-4}$

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