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Predicting Products/Balancing Equations Online Help

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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 CO2 + H2O
• 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: AlCl3 (aq) + Pb(NO3)2 (aq) –>  ?

1. Identify the type of reaction.

Double-replacement reaction.

You have 2 compounds made up of cations and anions.

AlCl3 (aq) + Pb(NO3)2 (aq) –>

2. Write out the products of the chemical reaction.

Switch the cations, since this is double-replacement:

AlCl3 + Pb(NO3)2 –> Pb Cl + Al NO3

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

Pb = +2, Cl = -1, so the formula is PbCl2.

Al = +3, NO3 = -1, so the formula is Al(NO3)3.

AlCl3 + Pb(NO3)2 –> PbCl2 + Al(NO3)3

3. Balance the equation.

AlCl3 + Pb(NO3)2 –> PbCl2 + Al(NO3)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.

2AlCl3 + Pb(NO3)2 –> 3PbCl2 + Al(NO3)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.

2AlCl3 + Pb(NO3)2 –> 3PbCl2 + 2Al(NO3)3

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

2AlCl3 + 3Pb(NO3)2 –> 3PbCl2 + 2Al(NO3)3

Last, check the nitrates (NO3), which now both have 3 on either side.

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

Answer to Problem: 2AlCl3 + 3Pb(NO3)2 –> 3PbCl2 + 2Al(NO3)3

Note: Printing this out might help.

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