Question

Problem PageQuestion 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. Clears your work. Undoes your last action. Provides information about entering answers.

Answer 1

Complete question

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Answer:

The value is  $K_c = 2.69 *10^{-5}$  

Explanation:

From the question we are told that

   The equation is  

           $Fe_20_3_{(s)}+3H_{(g)}\to2Fe_{(s)}+3H_2O_{(g)}$

Generally the equilibrium is mathematically represented as

        $K_c = \frac{[H_2O]^2}{[H_2]^3}$

Here $[H_2O]$ is the concentration of water vapor which is mathematically represented as

      $[H_2O ] = \frac{n_w}{V_s }$

Here $V_s$ is the volume of the solution given as 8.9 L

$n_w$ is the number of moles of water vapor which is mathematically represented as

        $n_w = \frac{m_w}{Z_w}$

Here  $m_w$  is the mass of water given as 2.00 g

and   $Z_w$  is the molar mass of water with value  18 g/mol

So  

         $n_w = \frac{2}{18}$

=>     $n_w = 0.11 \ mol$

So

     $[H_2O ] = \frac{0.11}{8.9 }$

=>   $[H_2O ] = 0.01236 \ M$

Also

$[H]$ is the concentration of hydrogen gas which is mathematically represented as

      $[H ] = \frac{n_v}{V_s }$

Here $V_s$ is the volume of the solution given as 8.9 L

$n_v$ is the number of moles of  hydrogen gas which is mathematically represented as

        $n_v = \frac{m_v}{Z_v}$

Here  $m_w$  is the mass of water given as 4.77 g

and   $Z_v$  is the molar mass of water with value  2 g/mol

So  

         $n_w = \frac{4.77}{2}$

=>     $n_w = 2.385 \ mol$

So

     $[H_2O ] = \frac{2.385}{8.9 }$

=>   $[H_2O ] = 0.265 \ M$

So

     $K_c = \frac{( 0.01236 )^3}{ (0.265 )^2}$

=>   $K_c = 2.69 *10^{-5}$