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Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem

belka [17]

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

Fe₂O₃ 3.95 g

H₂ 4.77 g

Fe 4.38 g

H₂O 2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

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 $1.0\times 10^{-4}$

6 0

3 years ago

Carbon hexafluroide formula

ikadub [295]

The formula for Hexafluoride is F6S.

4 0

3 years ago

When would formation of a derivative be a necessary step in this experiment?

const2013 [10]

The formation of a derivative being a necessary step in the experiment lies in the importance of the derived structure. Often the derived product confers to reaction pathways which uses less reactive starting materials and more easily proceeds to completion. This also allows us to take a small amount of sample. The derived product at times is a general compound allowing its easy analysis. Often we encounter a product but we find it difficult to analyse it in ways we want. Here lies the essence of forming a derivative which often are simpler compounds allowing easier analysis yet having similar functional groups and structural properties. Also sometimes we encounter problems when our desired product is unstable and forms stable degraded products. But if we somehow manage to synthesize a derivative it may be relatively stable and form no degradation products. It would be stable at least for a significant period of time making it easier to study its properties. The derived product also at times are synthesized using general reaction pathways facilitating a way of easier synthesis and helping it to correlate with other similar reaction pathways and products.So the above paragraph accounts for the need of derivatives. When we encounter problems similar to those mentioned above it becomes necessary for a researcher to form rather synthesize a derivative.

6 0

3 years ago

The table describes how some substances were formed.

gogolik [260]

Answer:

Formed by adding 5 g of sugar to 1 L of water

Explanation:

The substance formed by adding 5 g of suagr to 1 L of water is a mixture.

A mixture is defined as two or more substances that are physically mixed but do not react together. It thus means that the component substances of a mixture can be retrieved through appropriate means of separations.

A mixture of sugar and water can be separated by evaporation. The mixture is heated and the vapor from the boiling water can be condensed to retrieve the water while the residue will give the sugar that was initially dissolved in it.

The products formed from other descriptions in the illustration cannot be separated ordinarily and thus, are not considered to be mixture. The components have been chemically combined together.

6 0

2 years ago

NEED SOME HELP ASAP!!!!!!

Dimas [21]

I think D If not sorry for giving the wrong one

5 0

2 years ago