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 chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilbrium has the following composition compound amount Fe 3.95 g H 4.77 g Fe 4.38 g H2 2.00 g Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 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 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.

The value of equilibrium constant for given equation is

Explanation:

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

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:

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:

For the given chemical equation:

The expression of equilibrium constant for above equation follows:

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

Putting values in above expression, we get:

Hence, the value of equilibrium constant for given equation is

if you mean hydrogen and oxygen then water is the product