**Predicting Products/Balancing Equations
Online Help**

**Practice
Problems Here**:

Practice
Problem #3

Practice
Problem #4

**Here are the key steps (in order) to doing stoichiometry:**

**1.** Identify the type of reaction [combustion,
synthesis, decomposition, single-replacement, or double replacement].

**2.** Write the products of the reaction based on
reaction type.

**Combustion**: always produces CO_{2}+ H_{2}O**Synthesis**: combines the elements, then write the chemical formulas canceling out the charges**Decomposition**: breaks up the compounds into elements, being aware of the diatomic elements**Single-replacement**: cations take the place of cations, and anions take the place of anions; then write the chemical formulas canceling out the charges

· **double-replacement**: cations take the place of cations, and anions
take the place of anions; then write
the chemical formulas canceling out the charges

**3.** Then, balance the equation.

**Sample Example Problem [to show you how it's done]:**

**Problem:** AlCl_{3} (*aq*) + Pb(NO_{3})_{2}
(*aq*) --> ?

**1.**** Identify the type of reaction.**

Double-replacement reaction.

You
have 2 compounds made up of **cations** and **anions**.

AlCl_{3}
(*aq*) + Pb(NO_{3})_{2}
(*aq*) -->

**2.**** Write out the products of the
chemical reaction.**

Switch the **cations**,
since this is double-replacement:

AlCl_{3}
+ Pb(NO_{3})_{2}
--> Pb Cl + Al NO_{3}

Then write the formulas of the products based on their
charges.

Pb = +2, Cl = -1, so the formula is PbCl_{2}.

Al = +3, NO_{3} = -1, so the formula is Al(NO_{3})_{3}.

AlCl_{3}
+ Pb(NO_{3})_{2}
--> PbCl_{2}
+ Al(NO_{3})_{3}

**3.**** Balance the equation.**

AlCl_{3} + Pb(NO_{3})_{2} -->
PbCl_{2} + Al(NO_{3})_{3}

The number of Al
atoms is equal on both sides (both have 1).

Next, there are 3 Cl atoms on the left, 2 on the right.

Lowest common multiple is 6. See coefficients
below.

Both multiply the Cl atoms to 6.

**2**AlCl_{3}
+ Pb(NO_{3})_{2} --> **3**PbCl_{2}
+ Al(NO_{3})_{3}

Next, there is 1 Pb atom on the left, 3 on the right
(because of **3**x1).

2AlCl_{3} + **3**Pb(NO_{3})_{2}
--> 3PbCl_{2} + Al(NO_{3})_{3}

Last, check the nitrates (NO_{3}), which now both
have 3 on either side.

This is the same as checking the nitrogens and the
oxygens by themselves.

**Answer
to Problem:** **2AlCl _{3}
+ 3Pb(NO_{3})_{2} --> 3PbCl_{2} + Al(NO_{3})_{3}**

Note: Printing this out
might help.