Iron(III) oxide and hydrogen react to form iron and water, like this: (s)(g)(s)(g) At a certain temperature, a chemist finds that a reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition: compound amount Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

The question is incomplete, here is the complete question:

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

At a certain temperature, a chemist finds that a 5.4 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition:

Compound        Amount

            3.54 g

                  3.63 g

      Fe             2.37 g

                2.13 g

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

The value of equilibrium constant for the given reaction is

Explanation:

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

Given mass of hydrogen gas = 3.63 g

Molar mass of hydrogen gas = 2 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

Given mass of water = 2.13 g

Molar mass of water = 18 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

The given chemical equation follows:

The expression of for above equation follows:

The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression. So, the concentration of iron and iron (III) oxide is not present in equilibrium constant expression.

Putting values in above equation, we get:

Hence, the value of equilibrium constant for the given reaction is

it's c (engineer's do their work with mechanical stuff not on any living being)