Question

A common laboratory method for preparing oxygen gas involves decomposing potassium chlorate (CIO), as shown by the reaction. 2 KCIO, — 2 KCI + 302 Based on this equation, how many grams of KCIO, must be decomposed to produce 3.95 mol KCI? mass: g KCIO, How many grams of KCIO, must be decomposed to produce 89.7 g 0,? mass 8 KCIO,

Answer 1

Answer (a) As we can see that for every 2 moles of KClO₃ we get 2 moles of KCl so for 3.95 mol    

     KCl we need exactly 3.95 mol KClO₃.

     We have to calculate amount of KClO₃ in g i.e., mass = ?

      Molecular weight of KClO₃ = 122.55 g/mol

     As we know that

     No. of moles = Mass/Molecular weight

     3.95 = Mass/122.55

     Mass = 3.95 x 122.55

     Mass = 484.07g

(b) Similarly, For every 2 mol of KClO₃ we get 3 mol of O₂

         It means for every 0.66 mol KClO₃ we get 1 mol of O₂

        89.7g O₂ = 5.606 mol O₂

      Therefore for we need 5.606 x 0.66 g KClO₃ for 89.7g O₂ i.e., 3.699 mol

    3.699 mol KClO₃ = 453.31 g

     Hence, 453.31g KClO₃ must be decomposed to produce 89.7g O₂.