Predicting Reactions: A System

General Chemistry Index

Where are we going with this? This page will assist in developing the ability to predict products of simple reactions as listed in of reactions: synthesis (i.e., combination), decomposition, single displacement, double displacement, acid/base, and combustion.

Predicting Reactions: A System 
What happens if I mix this with that?

Predicting products of chemical reactions is a process by which potential reactants are scrutinized to determine if they will react and if so what product(s) will be formed.

What happens if I mix this baking soda with vinegar?
What happens if I let this spilled gasoline sit on the painted garage floor?
What happens if I pour bleach directly onto my clothes?

Predicting chemical reactions does not only take place in the lab, but actually is a part of everyday life! However, in the lab, we can be more specific and isolate the this and the that more.

So… I told you this would be long! Maybe another soda or cup of coffee?

Recall: the goal is to end up with neutral molecules no matter what you have to do to the subscripts on the PRODUCT side.

Let's just say we are starting with "real" molecules on the reactant side, to begin with, okay? Coming up with a system to predict the products is, at best, a starting place. 

Because of the vast scope of chemistry, there will be variations that this system will not cover (probably). And, there are always exceptions!

Back to the original question, but flipping it: If I mix this with that, what happens?

The system I'm going to offer expands on that question thusly:

What do you have to start with?

This step identifies what kind of reaction you could be looking at. Is it synthesis? Decomposition? Etc. 

 

What does it become? 

1. Will it even react? 
2. What combination of atoms are needed in the product such that the compound molecule is neutrally charged? (AKA what are the subscripts?) 
3. What is the FINAL balanced equation?

 

So, here is a "decision tree" for what to do to predict the products of a chemical reaction. 

Predicting the Products of Chemical Reactions Decision Tree

What do you have to start with?

• Two elements: goto Synthesis Reaction below (Click) 

• One compound: goto Decomposition Reaction below (Click)

• An element and a compound: goto Single Displacement Reaction below (Click) 

• Two compounds: goto Double Displacement Reaction below (Click) 

What does it become?

Synthesis Reaction (background)

 

First off, will they react? 

For two elements to react, they need to have opposite valence charges (one plus, one minus = one has four or less electrons in its valence orbitals, one has four or more)

Secondly, if they will react, how many of each are needed to get a neutrally charged compound molecule in the product?

The subscripts denote how many of each thing will be present in the neutrally charged molecule. You can "criss-cross" the charges of two elements (then reduce mathematically) to find the right numbers.

For instance, take carbon and oxygen.

Oxygen has a valence of -2.

Carbon has a charge of ±4. (Since oxygen is negative, we'll use the +4)

C + O2 --> ??

So, it has to be C?O?

Criss-cross the charges:  C2O4 

Reduce the subscripts mathematically: C2O4   becomes  CO2

C + O2 --> CO2

Thirdly, balance the equation such that the same number of each type of atom is present on both sides:

C + O2 --> CO2 (was already in balance)

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Decomposition Reaction (background)

 

First off, will they react? 

Not everything will break apart easily. Somethings only break apart at high temperatures.

It seems fair to presume that, if given a predicting products exercise, the compound will, by some means react and decompose into the parts.

Secondly, if… Well, this one is pretty easy. Whatever you start with breaks apart. But… into how many pieces! (Probably two.) Usually, it will look like this:

AB --> A + B 

Both the A and the B have to be neutral molecules or elements. And don't forget about those diatomic elements! 

Thirdly, balance the equation such that the same number of each type of atom is present on both sides:

2H2O → 2 H2 + O2

It could be tricky, though!

Na2CO3 → Na2O + CO2

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Single Displacement Reaction (background)

First off, will they react? 

For one thing to replace another… Let's say it like this… For A to replace B, 

A + BC --> ?? + ????

A has to be more highly reactive than B. How would anyone know that? There is a chart!

So, sodium won't replace potassium. Etc.!

 

Also, either B or C could be a polyatomic ion! That makes it harder to figure out what is being replaced. Look into the BC part and match one of them to the A with regards to location on the periodic table. There's a good chance that A will be in a family/group that is near the family or group of B or C. (You'll have to be openminded about this claim when dealing with transition elements.)

 

Secondly, if they will react, how many of each are needed to get a neutrally charged compound in the product?

You know what you are starting with, so the reactant side is done. Let's do aluminum and HCl as an example…

Al + HCl --> ?? + ?

Since Al will, in fact, replace H, the product side will be:

Al + HCl --> AlCl + H

Product side subscript time:

This should be fun!

1. H is diatomic, so it will be H2

2. The AlCl has to become neutrally charged. 

The subscripts denote how many of each thing will be present in the neutrally charged molecule. You can "criss-cross" the charges of two elements (then reduce mathematically) to find the right numbers.

For instance, take carbon and oxygen.

Al has a valence of +3.

Cl has a charge of -1.

The balanced molecule will be AlCl3

Thirdly, balance the equation such that the same number of each type of atom is present on both sides:

2Al + 6HCl --> 2AlCl3 + 3H2

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Double Displacement Reaction (background)

First off, will they react? 

For one thing to replace another… Let's say it like this… For A to replace B, 

AB + CD --> ???? + ????

Deciding what is A and what is B can be hard when polyatomic ions are involved. Really, the only way to get good at this is to do it a lot.

Look at A and C first. Are they both metals? Hmmm… What about B and D? Both polyatomic ions? If so, do they both have a positive or negative charge (There is a chart!)? If both B and D are negative and both A and C are positive (which is kinda the normal way of writing molecules) then you can find your potential "swaps."

At this point, you have to answer the question! Will they react?

Answering this question is complex, since the reactivity of each of the four parts is in play. How is it done? (CLICK HERE)

So… 

1. Does it form water? 

2. Does it form a gas? 

3. And then that insolubility thing… easy! 

Secondly, if they will react, how many of each part is needed to get a neutrally charged compound in the product?

The work done in the first step should have resulted in you knowing what the AB and CD parts are.  The product will become AD and BC

You know what you are starting with, so the reactant side is done. Let's do the follow as an example…

Fe2(SO4)3 + KOH → 

So, there's some AB and CD up there? That looks like a sentence in a foreign language!

So this… 

 

AB          +   CD  → AD + CB

Fe2(SO4)3 + KOH →

(When you predict the swapped product, put in the parentheses to start with, at least in your head.)

Fe2(SO4)3 + K(OH)1 →

Fe2(SO4)3 + K(OH)1 → Fe?(OH)? + K?(SO4)?

Product side subscript time:

This should be… never mind!

1. Do that thing with the charges to get balanced molecules. For instance…

K (from periodic table) has a charge of + 1

SO4 has a charge of -2

Criss-cross the charges to get K2(SO4)1

Chemistry "grammar" says we don't write ones and if the subscript is one don't use parentheses.

Thus, we get:

K2SO4 

Using the the example from above, working through the process of getting both products to a neutral charge, we get the unbalanced (but each part is a neutrally charged molecule) equation:

?Fe2(SO4)3 + ?KOH → ?K2SO4 + ?Fe(OH)3

Thirdly, balance the equation such that the same number of each type of atom is present on both sides.

So for the above example…

Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3