1. Word Problems: Predicting Products in Single and Double Replacement reactions
Given the beginning of a chemical reaction, how can you figure out what will be made, and if the reaction will actually happen?

2. Steps to follow when writing your own equations
Determine the type of reaction (Synthesis, Decomposition, Single Replacement, Double Replacement or combustion)
Use the reaction type information to predict the likely products
Check to make sure that each compound created has balanced charges (zero net charge) .

3. Steps to follow when writing your own equations – Continued
Determine if the reaction will happen
If Single Replacement:
A + BC  B + AC
Determine which elements will fight during the reaction. Two cations will fight over one anion. OR Two anions will fight over a single cation.
Zn + FeCl2  Fe + ZnCl2
F2 + NaBr  NaF + Br2
In the products, is the most reactive element bonded and the least reactive element alone?

4. Steps to follow when writing your own equations – Continued
Determine if the reaction will happen
If Double Replacement:
AB + CD  AD + CB
NaCl + AgNO3  NaNO3 + AgCl
Switch the partners in the compounds. Make sure that you bond the cation with a new anion. (The first ion bonds with the last ion and the two inner ions bond)
Is at least one of the products INSOLUBLE in water? Was a covalent substance (like water) produced? Was a gas created?

5. Solubility Table: s= soluble Ss= slightly soluble i=insoluble
This table came from

6. Steps to follow when writing your own equations – Continued
To sum up: Single replacement reactions occur if the most active metal of the two is bonded with a nonmetal OR the most active nonmetal of the two is bonded with a metal at the end of the reaction. Double replacement reactions occur if an insoluble ionic product, a gas or a covalent substance like water is formed. If it is any other type of reaction (synthesis, decomp or oxidation) you may assume that it proceeds as written. If the reaction will occur, then use coefficients to balance it. If the reaction does not occur, then you can skip the balancing step.

7. Iron nails are placed in a solution of copper (II) chloride…
Example Problem 1
Iron nails are placed in a solution of copper (II) chloride…
Fe + CuCl2 
Looks like Single Replacement
A + BC  AC + B
(Fe replaces the Cu+2 ion in the solution)
Fe + CuCl2  FeCl + Cu
Iron is more active than copper, so it should be able to replace it in solution.
Iron takes either Fe+2 or Fe+3 charge. To make it easy, let’s assume Fe+2 is created.
____Fe + ____CuCl2  _____FeCl2 + ____Cu
(Given more time, we could figure out which compound is actually created by doing some stoichiometry and/or looking at the color of the solution formed.)

8. Lead (II) nitrate and potassium iodide solutions are mixed…
Example Problem 2
Lead (II) nitrate and potassium iodide solutions are mixed…
Pb(NO3)2 + KI 
Looks like Double Replacement
AB + CD  AD + CB
Lead will drop the nitrate and try to bond with the iodine. Potassium will also switch partners
Pb(NO3)2 + KI  PbI2 + KNO3
PbI2 = INSOLUBLE KNO3 = Soluble
Since one of the products is insoluble, the reaction will occur
___Pb(NO3)2 + ___KI  ___PbI2 + ___KNO3
(Given more time, we could figure out which compound is actually created by doing some stoichiometry and/or looking at the color of the solution formed.)

9. Potassium iodide and sodium carbonate solutions are mixed…
Example Problem 3
Potassium iodide and sodium carbonate solutions are mixed…
KI + Na2CO3
Looks like Double Replacement
AB + CD  AD + CB
Potassium will drop the iodide ion and try to bond with the carbonate. Sodium will also switch partners
KI + Na2CO3 NaI + K2CO3
Both products are soluble ionic compounds. No gas was created. No reaction occurred.
There is no need to balance it. The reaction does not occur.
(Given more time, we could figure out which compound is actually created by doing some stoichiometry and/or looking at the color of the solution formed.)

10. Example Problem 4 NaI + Cl2  Looks like Single Replacement
A solution of sodium iodide is mixed with chlorine gas dissolved in water…
NaI + Cl2 
Looks like Single Replacement
AB + C  AC + B
(Chlorine will attempt to replace the iodine)
NaI + Cl2  NaCl + I2
Chlorine is more active than iodine so the reaction will proceed.
____NaI + _____Cl2  _____NaCl + _____I2
(Given more time, we could figure out which compound is actually created by doing some stoichiometry and/or looking at the color of the solution formed.)

11. Try these two problems on your own
Practice Problem A
Barium nitrate and sodium carbonate solutions are mixed…
Practice Problem B
A copper metal strip is placed in sulfuric acid…
Pause the video and try these two problems on your own