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 5.4 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition: Calculate the value of the equilibrium constant K_c for this reaction. Round your answer to 2 significant digits.

Answer 1

Complete Question

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Answer:

The equilibrium constant is  $K_c= 2.8*10^{-4}$

Explanation:

      From  the question we are told that

              The chemical reaction equation is

      $Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}$

The voume of the misture is  $V_m = 5.4L$  

  The molar mass of  $Fe_{2} O_{3}_{(s)}$ is a constant with value of  $M_{Fe_{2} O_{3}_{(s)} } = 160g/mol$

    The molar mass of  $H_{2}_{(g)}$    is a constant with value of  $H_2 = 2g/mol$

   

    The molar mass of  $H_{2}O$    is a constant with value of  $H_2O = 18g/mol$

Generally the number of moles  is mathematically given as

                     $No \ of \ moles \ = \frac{mass}{molar\ mass}$

    For   $Fe_{2} O_{3}_{(s)}$

          $No \ of\ moles = \frac{3.54}{160}$

                                $= 0.022125 \ mols$

     For  $H_{2}$

               $No \ of\ moles = \frac{3.63}{2}$

                                $= 1.815 \ mols$

       For  $H_{2}O$

                         $No \ of\ moles = \frac{2.13}{18}$

                                              $= 0.12 \ mols$

Generally the concentration of a compound  is mathematicallyrepresented  as

       $Concentration = \frac{No \ of \ moles }{Volume }$

      For   $Fe_{2} O_{3}_{(s)}$

                $Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}$

                                         $= 4.10*10^{-3}M$                          

       For  $H_{2}$

                  $Concentration[H_2] = \frac{1.815}{5.4}$

                                           $= 0.336M$

      For  $H_{2}O$

                $Concentration [H_2O] = \frac{0.12}{5.4}$

                                                  $= 0.022M$

  The equilibrium constant  is mathematically represented as

                $K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}$

  Considering $H_2O \ for \ product$

            And      $H_2 \ for \ reactant$

At  equilibrium the

                    $K_c = \frac{0.022}{0.336}$

                          $K_c= 2.8*10^{-4}$