CHM 2045C Module Five Sample  Exam     

 

Module Five (Question #70 p 171 Kotz 6th)

 

Part N  Multi-step Synthesis Stoichiometry       4  points

 

Potassium perchlorate may be made by the following series of reactions:

 

                Cl2   +   2 KOH      ---->    KCl    +    KClO      +     H2O

 

                           3 KClO        ----->      2 KCl    +      KClO3

 

                           4 KClO3      ----->      3   KClO4        +        KCl

 

How much Cl2 is needed to prepare 100 g of Potassium perchlorate KClO4 by the above sequence?

 

____?_____ g Cl2  =  100 g KClO4 

 

  1. Change 100 grams KClO4 to moles of KClO4 using molar mass of KClO4

 

  1. Then apply three mole ratios from the three equations working your way backward from equation #3 to the Cl2 in Equation #1:

 

 

3 moles KClO4 = 4 moles KClO3

1 mole KClO3 = 3 mole KCLO

1 mol KClO = 1 mole Cl2

 

  1. Then change the moles of Cl2 back to grams by using molar mass of Cl2

 

 

 

 

 

 

Other Problems: Problem #73 Page 171 Kotz 6th Edition