If 5.0 g of potassium chlorate (KClO3) is decomposed, what volume of oxygen gas is produced at STP?

1.37dm³

Explanation:

To solve this problem, let us write the reaction expression:

               2KClO₃  →   2KCl  + 3O₂

Now, mass of KClO₃ is 5g, let us find the number of moles;

  Number of moles  =  

  Molar mass of KClO₃  = 39 + 35.5 + 3(16) = 122.5g/mol

 Now;

 Number of moles  =   = 0.04mole

  So;

       2 moles of KClO₃  will produce 3 moles of oxygen gas

       0.04 mole of KClO₃  will produce   = 0.06moles

At STP;

        1 mole of gas has  a volume of 22.4dm³

       0.06 mole of oxygen gas will have a volume of 22.4 x 0.06 = 1.37dm³

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