Tuesday, May 17, 2011

Exam Review #16

16)    Predict the products of the following reactions and write balance chemical equations:
a.     The incomplete combustion of propane (C3H8
b.     Solutions of potassium bromide and lead (II) nitrate are mixed. (include net ionic equation)
c.     Chlorine gas is bubbled into a solution of potassium iodide ( include net ionic equation)
d.     Dinitrogen trioxide gas is passed over water
e.     Solid copper (II) oxide is added to water
f.      The reaction of nitric acid with potassium hydroxide solutions (include net ionic equation)
g.     Aluminum phosphate is strongly heated
h.     A piece of silver metal is added to a solution of aluminum nitrate
i.      Solid barium chlorate is heated in the presence of magnesium dioxide as a catalyst.
j.      Solutions of magnesium sulfate and copper nitrate are mixed (include net ionic equation) 


  1. Unfortunately, I completely disagree with (g) and (i). I think you may have misunderstood how a decomposition reaction works. It would be too simple just to break the compound up into its ions. Both of these required you to use oxidation numbers. For (g) I get aluminum oxide and diphosphorous pentoxide as products. For (i) I get barium oxide and dichlorine trioxide.

  2. Cyrus, I have reviewed these two problems and have come up with new answers... see above. I agree with you on (g), but I believe that your correction for my answer on (I) is wrong... I am getting BaCl2 and 3-(O)2

  3. I tentatively agree. I believe that the decomposition of a metal halate may have the same pattern as that of a halite. However, I am sure of the fact that it wouldn't be 3O2, it would be 2O2.

  4. Good job on your review. It was really helpful that you posted all your work, but it says on question C that the reaction does not occur because of K. I got that the reaction does occur because chlorine is above iodine on the scale.

  5. I agree with michelle, it does not matter that K is higher than Cl. Cl is replacing the non metal iodine, not K.

  6. And I think it should be 3 O2 on i as well..