**Predicting Products/Balancing Equations Online Help**

**Practice Problems Here:**

Practice Problem #1

Practice Problem #2

Practice Problem #3

Practice Problem #4

Practice Problem #5

**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, the are 2 Al atoms on the left, 1 on the right.

Lowest common multiple is 2. See coefficients below.

Both multiply the Al atoms to total 2 atoms Al on either side.

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

Note: Printing this out might help.

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