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

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current