Reactions The study of Chemistry would be incomplete without an understanding of reactions. To fully grasp reactions, you should understand what a reaction is (also called a chemical change) and what makes it different from a physical change. You should be able to recognize the five basic types of reactions and you should be able to complete each type (figure out the products from the reactants), balance reactions, and you should be able to determine whether or not single displacement reactions occur if given an activity series and you should be able to determine whether or not double displacement reactions occur given a solubility table.
Writing reactions When a reaction occurs, we can record on paper what has happened, but we do so in very specific ways. For example, when methane (the gas that is probably used in your chem lab burners, or in your home kitchen – formula CH4) is burned it is reacting with oxygen. The reaction that occurs is written this way:
image – rxn1 (shows balanced reaction and labels reactants, products, subscripts, coefficients)
When reactions are written in this manner, there are a few very important rules that must be followed: 1. The reactants (the things that are reacting together) are always on the left side 2. An arrow leads from the left to the right and signifies that a change is occurring. 3. The products (the things that are formed in the reaction) are always on the right side 4. The reaction must be balanced. That means that there must be the same number of atoms of each element on both sides of the reaction. 5. Formulas MUST be written correctly according to the defined practice 6. Reactions are balanced by adding coefficients (by changing how many or each molecule are involved in the reaction.
Here is the same example with explanations image – rxn2 ~~~~~~~~~~~~~~~~ Balancing reactions We'll start balancing reactions assuming that you already have the formulas written correctly. This is very important since it is often impossible to balance a reaction in which the formulas are written incorrectly, and even if you get it balanced, it is still wrong. If you aren’t sure about writing formulas, you can go back to nomenclature.
Balancing reactions is largely a process of trial and error. There is, unfortunately, no simple process that always works. There are a few common situations and types that we will try to show here, but in the end, you will have to do some guess and check work.
Let's start with a simple example: Ba(NO3)2 + Na2SO4 → NaNO3 + BaSO4
If we do a quick count we discover that there are the following atoms on each side of the equation: Left side Right side Ba = 1 Ba = 1 N = 2 N = 1 Na = 2 Na = 1 O = 10 (6 from the nitrates and 4 from sulfate) O = 7 S = 1 S = 1
It is also perfectly okay (and often easier) to count polyatomic ions as complete units. Of course this only works if the ion is identical on both sides. In other words, it will work here, but it wouldn't work in an equation that had sulfate on one side and sulfite on the other.
If we count by ions, we would have: Ba(NO3)2 + Na2SO4 → NaNO3 + BaSO4 Left side Right side Ba = 1 Ba = 1 NO3 = 2 NO3 = 1 Na = 2 Na = 1 SO4 = 1 SO4 = 1
The reaction is not balanced now because there are different numbers of nitrates and sodium atoms. So what can we do? ~~~~~~~~~~~~~~~~~~~~~~ Here’s that same unbalanced reaction Ba(NO3)2 + Na2SO4 → NaNO3 + BaSO4 Left side Right side Ba = 1 Ba = 1 NO3 = 2 NO3 = 1 Na = 2 Na = 1 SO4 = 1 SO4 = 1
The reaction is not balanced now because there are different numbers of nitrates and sodium atoms. We are NOT allowed to alter the formulas for the compounds. In other words we cannot change the Na2SO4 into NaSO4 so, instead, we need to add coefficients.
One good way to start is to begin at the left. The first thing we run into is Ba, however barium is balanced (1 on each side). The next thing we run into is nitrate. This is unbalanced (2 on the left and only 1 on the right). So, I need another nitrate on the right, but remember, I cannot change the formula, so I'll add a coefficient of 2 in front of the sodium nitrate.
Ba(NO3)2 + Na2SO4 → 2 NaNO3 + BaSO4
Now if I count again:
Left side Right side Ba = 1 Ba = 1 NO3 = 2 NO3 = 2 Na = 2 Na = 2 SO4 = 1 SO4 = 1
As you can see, everything is balanced, so we're done.
Let's try another example ~~~~~~~~~~~~~~~~ Balancing Reactions – another example Let's try another example of balancing reactions. Hg(NO3)2 + AlCl3 → Al(NO3)3 + HgCl2 If we count elements and ions, we find: Left side Right side Hg = 1 Hg = 1 NO3 = 2 NO3 = 3 Al = 1 Al = 1 Cl = 3 Cl = 2
This is obviously not balanced, so let's see what we can do. Starting at the left, we see that the mercury's are balanced, but the nitrates are not. We have three nitrates on the left and two on the right. There is, unfortunately, no single number I can put in as a coefficient to make these match. There is, however a relatively easy solution. The nitrates on the left come in pairs. That means that I must always have 2 or 4 or 6, or 8, etc. nitrates on the left side as I increase the coefficient. On the right side, the nitrates come in trios. That means that as I add coefficients, I will have 3 or 6 or 9, etc. nitrates. The connection? 6. In other words I need to get 6 nitrates on each side to balance them.
Another way of arriving at this number (for those who are mathematically minded) is to look for the least common multiple (the lowest number that each original value can be multiplied to form). The least common multiple of 2 and 3 is 6. That means that I need to have 6 nitrates on both sides to balance the equation.
A note on 2’s and 3’s
To achieve these matching 6's, I need a coefficient of 3 on the left and a coefficient of 2 on the right: 3 Hg(NO3)2 + AlCl3 → 2 Al(NO3)3 + HgCl2 Now if I count elements and ions, I get:
Left side Right side Hg = 3 Hg = 1 NO3 = 6 NO3 = 6 Al = 1 Al = 2 Cl = 3 Cl = 2 Now my nitrates match, but my mercury's no longer do. However, I can solve that by adding a coefficient of 3 to the mercury compound on the right: 3 Hg(NO3)2 + AlCl3 → 2 Al(NO3)3 + 3 HgCl2
Left side Right side Hg = 3 Hg = 3 NO3 = 6 NO3 = 6 Al = 1 Al = 2 Cl = 3 Cl = 6 Getting closer. Now both the nitrates and the mercury's match, but the aluminum's and chlorides are still a problem. Focusing just on the aluminum (1 on the left and 2 on the right) it seems clear that I should add a coefficient of 2 on the left side before the aluminum compound. 3 Hg(NO3)2 + 2 AlCl3 → 2 Al(NO3)3 + 3 HgCl2
Left side Right side Hg = 3 Hg = 3 NO3 = 6 NO3 = 6 Al = 2 Al = 2 Cl = 6 Cl = 6 So, now my mercury's, nitrates and aluminum's all match and...wait a minute...the chlorides do as well. The reaction is balanced. A few notes on balancing ~~~~~~~~~~~~~~ A few notes on balancing. It is NOT required to make a table, as I have on these pages, to count and it is certainly not required to make multiple tables. I promise you that your chemistry teacher does not make tables when they are balancing reactions on their own. It is perfectly okay to do all of this in another form (or in your head) as long as you end up with the correct balanced reaction.
A note on 2’s and 3’s and other non-multiples The pattern that we saw in the last reaction (2 of something on one side and 3 on the other) is VERY common and the solution is ALWAYS the same. The side with 2 is multiplied by 3 and the side with 3 is multiplied by 2. In “lazy speak” – when you have a 2 and 3, just 3 and 2 it.
Of course this idea works with any pair of numbers. If you have a 5 and 4, just 4 and 5 it.
But beware, if you have a 2 and 4 you do NOT want to 4 and 2 it. You just want to double the 2. We are always looking for the simplest whole number ratios.
There are a few other common patterns that are associated with particular types of reactions. Those will be dealt with later.
Let's look at a third example: HCl + Ca(OH)2 → H2O + CaCl2 In this reaction we cannot count the ions (hydroxide) because there is no hydroxide on the right side. Therefore we will have to deal with each element separately. Left Side Right Side H = 3 H = 2 Cl = 1 Cl = 2 Ca = 1 Ca = 1 O = 2 O = 1 The Ca's are balanced, but nothing else is. Although it is tempting to start with the H's (since they appear first) it will be easier to NOT do this because the H's appear in BOTH of the compounds on the left side. This means that there are two different ways that the number of hydrogen atoms can be changed. Instead, let's skip the H's and move on to the Cl's. There is 1 Cl atom on the left and 2 on the right. So, we'll add a coefficient of 2 on the left side and then recount: 2 HCl + Ca(OH)2 → H2O + CaCl2
Left Side Right Side H = 4 H = 2 Cl = 2 Cl = 2 Ca = 1 Ca = 1 O = 2 O = 1 Now, the calcium and chlorine are both balanced. Still avoiding the hydrogen, we can look at the oxygen (2 on the left and one on the right). Adding a coefficient of 2 on the right side (for the water) gives: 2 HCl + Ca(OH)2 → 2 H2O + CaCl2
Left Side Right Side H = 4 H = 4 Cl = 2 Cl = 2 Ca = 1 Ca = 1 O = 2 O = 2 Everything is balanced.
So, a general rule: It is possible to start balancing by looking at any element or ion in the reaction, but if an element appears in more than one place on the same side, it may be worth avoiding.
One last example: Na3PO4 + BaCl2→ Ba3(PO4)2 + NaCl Counting the elements and ions we get: Left side Right side Na = 3 Na = 1 PO4 = 1 PO4 = 2 Ba = 1 Ba = 3 Cl = 2 Cl = 1 If we start at the left with Na... Na3PO4 + BaCl2→ Ba3(PO4)2 + 3 NaCl Left side Right side Na = 3 Na = 3 PO4 = 1 PO4 = 2 Ba = 1 Ba = 3 Cl = 2 Cl = 3 But now, if we move on to the phosphates...we end up UNbalancing the sodium's 2 Na3PO4 + BaCl2→ Ba3(PO4)2 + 3 NaCl Left side Right side Na = 6 Na = 3 PO4 = 2 PO4 = 2 Ba = 1 Ba = 3 Cl = 2 Cl = 3
At this point, you may see the solution, but if not, it might be worth going back to the beginning and starting with a different element or ion. Na3PO4 + BaCl2→ Ba3(PO4)2 + NaCl Counting the elements and ions we get: Left side Right side Na = 3 Na = 1 PO4 = 1 PO4 = 2 Ba = 1 Ba = 3 Cl = 2 Cl = 1 If we take a quick look at this reaction, we see that the barium phosphate on the right side is the most complex formula in the problem – that is, this is the compound with the most subscripts in it. Let's start there, with the barium (although it would work just as well to start with the phosphates. Na3PO4 + 3 BaCl2→ Ba3(PO4)2 + NaCl Left side Right side Na = 3 Na = 1 PO4 = 1 PO4 = 2 Ba = 3 Ba = 3 Cl = 6 Cl = 1 Since this changed the value for chloride, lets look at that next... Na3PO4 + 3 BaCl2→ Ba3(PO4)2 + 6 NaCl Left side Right side Na = 3 Na = 6 PO4 = 1 PO4 = 2 Ba = 3 Ba = 3 Cl = 6 Cl = 6 We can now look at the phosphates or the sodium's to finish the equation 2 Na3PO4 + 3 BaCl2→ Ba3(PO4)2 + 6 NaCl ...and everything is balanced.
Try these reactions on your own. Pb(C2H3O2)2 + KBr → KC2H3O2 + PbBr2
AlCl3 + LiOH → LiCl + Al(OH)3
Cu(NO2)2 + FeCl3 → CuCl2 + Fe(NO2)3
The Five Basic Reactions Types There are five common types of reactions that you will be dealing with here. (There are more, notably redox reactions and organic reactions) but these five will get you pretty far. The five types of reactions are Double Displacement reactions Single Displacement reactions Combustion reactions Synthesis reactions Decomposition reactions
For each type of reaction, you must be able to recognize the reaction type, you must be able to complete the reaction (that is you must be able to determine the products of the reaction if you are given the reactants), and you must be able to balance them. In addition, for both double displacement reactions and single displacement reactions you should be able to determine (using the appropriate resource) whether or not the reaction will occur in the lab.
Double Displacement reactions Double displacement reactions (also known as double replacement reactions) involve the swapping of ions between two ionic compounds or between one ionic compound and one acid. Recognizing Double Displacement Reactions So, a double displacement reaction is a reaction that starts with two ionic compounds or one ionic compound and one acid. Completing Double Displacement Reactions The products of a DD reaction are the result of the ions trading partners. In simpler terms, each positive ion starts with a negative ion as a “partner.” At the end of the reaction, the positive ions are paired with the “other” negative ion. Here's an example: NaCl + AgNO3 → NaNO3 + AgCl In this reaction, the sodium started the reaction with the chloride and ended with the nitrate. At the same time the silver started with the nitrate and ended with the chloride. Here is another example: Li2SO4 + BaCl2 → LiCl + BaSO4 In this (currently unbalanced) reaction, the barium started with the chloride and ended the reaction with the sulfate, while the lithium started with the sulfate and ended with the chloride. Take a close look at this reaction. You will notice that there are 2 lithium ions in the formula on the left, but only one on the right. The reason is very straightforward. The charge on lithium ions is +1, while the charge on sulfate ions is -2. Therefore when lithium ions are paired with sulfate, there must be two of them to balance the charge. However, the charge on chloride is -1, so when lithium and chloride are paired, only one of each is needed. This is the MOST LIKELY place to make a MISTAKE!. Formulas MUST be written according to the charges of the ions, not based on the number of the ions present on the other side of the reaction. Of course, we must deal with this inequality, which is why we balance the reaction. Li2SO4 + BaCl2 → 2 LiCl + BaSO4
Here are some additional examples: Al(NO3)3 + KOH →
Rearranging the ions gives: Al(NO3)3 + KOH → AlOH + KNO3
Writing the formulas on the product side of the reaction (the right) correctly (taking into account the charges on the ions), we get: Al(NO3)3 + KOH → Al(OH)3 + KNO3
Then, balancing the equation gives us: Al(NO3)3 + 3 KOH → Al(OH)3 + 3 KNO3
Another example: H3PO4 + Pb(C2H3O2)2 →
Switching the ions, gives us: H3PO4 + Pb(C2H3O2)2 → PbPO4 + HC2H3O2
Taking into account the ion charges on the right side of the equation: H3PO4 + Pb(C2H3O2)2 → Pb3(PO4)2 + HC2H3O2
Note: in the work above, there were two possible charges for lead. The key to getting it right is that charge NEVER changes in double displacement reactions. So, since the Pb is a +2 ion on the left, it must still be a +2 ion on the right.
Balancing the reaction gives us: 2 H2PO4 + 3 Pb(C2H3O2)2 → Pb3(PO4)2 + 6 HC2H3O2
Single Displacement Reactions Single displacement reactions (also known as single replacement reactions) occur between an element and an ionic compound or between and element and an acid. In almost all cases the element is either a metal or hydrogen gas (which acts like a metal since it has a positive charge when it is an ion). The rare exceptions to this rule are treated here. Here: Special single displacement reactions On rare occasions, single displacement reactions occur between a halogen and an ionic compound (which contains a halogen as the negative ion). In these cases, the halogens trade places. For instance:
2 AlBr3 + 3 F2 → 2 AlF3 + 3 Br2
Do they occur? The activity series for halogens is the periodic table. This reaction occurs because fluorine is more active than bromine.
Recognizing Single Displacement Reactions A single displacement reaction starts with an element and a compound. The element is usually a metal, but it can also be hydrogen (H2) or a halogen (F2, Cl2, Br2, I2)
Completing Single Displacement Reactions Single displacement reactions involve trading partners just like double displacement reactions, but here one element starts alone and therefore one element will end alone. Here is a simple example:
Na + AgNO3 → NaNO3 + Ag
In this case the sodium joins up with the nitrate ion and the silver ends up alone.
Of course, we need to be careful about charges, just as we did with double displacement reactions. For example:
Na + SrCl2 → NaCl + Sr
This reaction is NOT balanced, but it is important to note that there are two chlorides on the left to match the charge of Sr (+2) but only one chloride on the right to match the charge (+1) of the sodium.
We deal with the mismatch of chlorides by balancing:
2 Na + SrCl2 → 2 NaCl + Sr
Here are a few other examples:
Ca + Al(OH)3 →
In this reaction, the calcium will replace the aluminum and connect to the hydroxide
Ca + Al(OH)3 → Ca(OH) + Al
Once we have determined which elements and ions are paired up, we need to consider charge. The charge on calcium is +2, while the charge on hydroxide is -1, so we will need 2 hydroxides to make that formula work.
Ca + Al(OH)3 → Ca(OH)2 + Al Of course we still need to balance the reaction, but it is VERY important to make sure that you have written the formulas correctly BEFORE you balance the reaction.
3 Ca + 2 Al(OH)3 → 3 Ca(OH)2 + 2 Al
This becomes a little more complicated when the element on the left can have more than one charge in a compound, as the following example shows:
Cu + AgNO3 → CuNO3 + Ag
We've paired up the copper with the nitrate and kicked out the Ag (as we should have), but now we need to figure out the correct formula for the copper nitrate. Copper has two possible charges (+1 and +2). As a general rule in single displacement reactions, the more common charge is the one used. In this case, +2 is the more common charge for copper. That gives the formula Cu(NO3)2, and the reaction becomes:
Cu + AgNO3 → Cu(NO3)2 + Ag
After balancing, we have:
Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag
General Rule: In single displacement reactions, always use the more common charge.
Remember: hydrogen is a “HOFBrINCl.” In other words, it always appears as H2 not as H
2 HCl + Zn → H2 + ZnCl2
Try these on your own:
Ba + CuBr2 →
Al + AgNO3 →
Co + AuF3 →
H3PO4 + Sr →
Do they occur? Not all single displacement reactions that can be written will actually happen in the real world. For instance, look at the two reactions below
Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag
2 Ag + Cu(NO3)2 → 2 AgNO3 + Cu
These are (obviously) the same reaction but backwards. In the first reaction copper is replacing silver in the compound and in the second, silver is replacing copper in the compound. Logically, only one of those is going to happen. Either nature prefers to have copper in compounds (in which case the first reaction occurs) or nature prefers to have silver in the reaction (in which case the second one occurs).
Another way to think about this is that if both reactions occur in nature than this mixture would go on reacting forever – the copper kicking out the silver and then the silver kicking out the copper, on and on. We know from experience that reactions don't go forever, so nature clearly favors one over the other.
So, how can we tell which reaction nature prefers? The answer is a property called activity.
An active element is one that is likely to be found in a compound. In inactive element is one that is likely to be found in pure form. Let's apply that idea to the example above. Copper is known to be more active (more likely to be in a compound) than silver. So the first reaction
Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag makes sense. The more active copper ends up in the compound.
The second reaction
2 Ag + Cu(NO3)2 → 2 AgNO3 + Cu doesn't make sense. The more active metal (the one more likely to be in a compound) is leaving the compound.
So, we can say that the first reaction (Cu + AgNO3 → ) would occur, because the more active metal goes into the compound, while the second reaction (Ag + Cu(NO3)2 → ) would not occur, because the more active metal is already in the compound.
Two more examples:
It is know that iron is more active than gold, so the reaction
Fe + AuCl3 → would occur, because that would put the more active metal (iron) in the compound.
It is also known that hydrogen is more active than Pt, so the reaction
HCl + Pt → will NOT occur, because the more active metal is already in the compound.
How do you (as a student) know which element is more active? It's actually quite simple. There is a list of metals (and hydrogen) from most active to least active called an activity series. You can find one here. Use the chart to determine whether each of the following reactions will occur or not: H2SO4 + Ca → Cu + CoCl2 → K + NaCl → H2 + Fe3(PO4)2 →
Do they occur? (DD) Often when solutions of two ionic compounds are mixed a precipitate (solid) is formed. The mixture may turn cloudy, or may even separate, with the solid settling at the bottom of the container. Occasionally, the mixture may bubble as a gas is produced. Other times however, nothing happens and the two solutions will mix and remain clear.
Predicting whether a double displacement reaction will occur is based on the following rule:
Rule: Double displacement reactions occur if ONE of the products is water, a gas or insoluble (a solid).
The first part of the rule is easy. The following reaction occurs because one of the products is water: H2SO4 + 2 NaOH → Na2SO4 + 2 H2O (If you were just switching partners you might have ended up with HOH. That is absolutely okay and is exactly the same thing as H2O. In fact, it's easier to balance if you write the product as HOH.)
To use the second part of the rule, you need to know the four common gases that are produced in double displacement reactions. They are: H2S, H2CO3, H2SO3 and NH4OH. So the following reaction occurs because one of its products is a gas:
HCl + Na2S → NaCl + H2S
A complication that your instructor may or may not care about... Three of the gases mentioned above are not actually gases. They are unstable compounds that immediately decompose to form gases. H2CO3 → H2O + CO2 (gas) H2SO3 → H2O + SO2 (gas) NH4OH → H2O + NH3 (gas)
That means that the reaction between HCl and Li2CO3 could be written 2 ways: 2 HCl + Li2CO3 → 2 LiCl + H2CO3 or 2 HCl + Li2CO3 → 2 LiCl + H2O + CO2
The second is more complicated, but also more honest.
The third part of the rule for double displacement reactions is also simple, if you know what things are soluble and what things are insoluble. There are two ways to accomplish this. Your instructor may give you a series of rules about solubility to learn. (One such set of rules can be found here.) Or your instructor may have you use a solubility table, like the one found here. On the table, S is for soluble and I is for insoluble. So the last part of the rule could be that double displacement reactions occur if one of the products is an I on the solubility table.
For example, the reaction between Na2SO4 and Ba(NO3)2 occurs because one of the products (BaSO4) is insoluble. Na2SO4 + Ba(NO3)2 → BaSO4 + 2 NaNO3 Combustion Reactions Combustion reactions occur between oxygen and almost anything else. These are the reactions that we think of as “fire” but they can occur so slowly we barely feel any warmth if at all and so quickly that they explode and then are finished.
Recognizing Combustion Reactions Combustion reactions are easy to spot they always have 2 reactants, one of which is O2. (Remember that oxygen is one of the HOFBrINCl's and therefore appears as O2 not O.)
Completing Combustion Reactions Most combustion reactions (at least those that occur in first year chemistry) involve hydrocarbons and a few other simple organic compounds. What that means is that probably everything your teacher burns (for real or on paper) will be made of carbon and hydrogen and perhaps some oxygen.
Competing these reactions is very straightforward. If the reactant includes hydrogen, the product will be water (H2O). If the reactant contains carbon, the product will be carbon dioxide (CO2). If the reactant contains both, then there will be two products (H2O and CO2). Extra oxygen in the reactant will NOT change the products. Here are a few examples (These reactions are NOT balanced, so that you can focus on the the identity of the products.) C + O2 → CO2 H2 + O2 → H2O CH4 + O2 → CO2 + H2O C2H5OH + O2 → CO2 + H2O C12H22O11 + O2 → CO2 + H2O
Balancing combustion reactions is not really different from balancing other types of reactions, but there are a few common patterns that you should recognize. We'll look at three reactions.
CH4 + O2 → CO2 + H2O
The first helpful thought is that you should balance the oxygen last. This is not because oxygen is somehow special, it is because it is alone on the left. In other words, oxygen is the only element in this reaction that can be balanced (with a coefficient) without changing at least two different element amounts. (If you put a coefficient in front of CH4, you alter both C and H, etc.)
In this case there is one C on the left and one on the right...so far so good.
There are 4 H atoms on the left, but two on the right. That can be fixed easily:
CH4 + O2 → CO2 + 2 H2O
Now both C and H are balanced. There are 2 O atoms on the left, and 4 on the right (2 in the CO2 and 2 in the 2 water molecules). Again, an easy fix.
CH4 + 2 O2 → CO2 + 2 H2O
Lots of combustion reactions work this neatly. Some however are more difficult:
C2H6 + O2 → CO2 + H2O
Two carbon atoms on the left and one on the right – that can be easily fixed.
C2H6 + O2 → 2 CO2 + H2O
There are 6 H atoms on the left and 2 on the right. This, also, can be easily fixed.
C2H6 + O2 → 2 CO2 + 3 H2O
Now, however, we have a problem. There are 7 oxygen atoms on the right (4 from the two CO2 molecules and 3 more from the water), but the oxygen comes in pairs (O2 on the left).
There are LOTS of ways to think your way out of this, but the easiest is the simple advice “When in trouble...double.” In other words, when you find an odd/even issue, try doubling everything (at least everything that you've already dealt with).
2 C2H6 + O2 → 4 CO2 + 6 H2O
Now if we re-count the oxygen atoms we have 14 on the right (8 from the CO2's and 6 from the waters). This is easy to work out.
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
A last example of combustion balancing: C2H5OH + O2 → CO2 + H2O
We can start with the C's and quickly get to this point:
C2H5OH + O2 → 2 CO2 + H2O
The first stumbling block is to mis-count the H atoms on the left side of the reaction. There are 6, not 5. Even though they are written separately, they all count together. That leads us to:
C2H5OH + O2 → 2 CO2 + 3 H2O
If we count oxygen on the right side we get 7 (just like last time) and so you might be tempted to think that this reaction needs to be doubled. But...oxygen is in BOTH reactants, not just the O2. So, ONE of the O atoms comes from the C2H5OH, the rest of them (the other 6 needed to make 7) must come from the O2. That means that we need 3 oxygens:
C2H5OH + 3 O2 → 2 CO2 + 3 H2O
Synthesis reactions Synthesis reactions are reactions in which 2 or more reactants join together in a single product.
Recognizing synthesis reactions Synthesis reactions are easy to spot if you are given both sides of the reaction (reactants and products) – quite simply, there is only one product. That means, of course, one compound, not that the coefficient is one.
2 Na + Br2 → 2 NaBr
2 NH3 + H2O + CO2 → (NH4)2CO3
Although there are many different synthesis reactions, you will only be able to predict the products (at least at this point in your chemistry career) if the reactants are elements. So, a reaction in which the only reactants are elements, must be a synthesis reaction.
Ca + H2 →
Fe + S8 →
Completing synthesis reactions Simple synthesis reactions (the only ones you'll be completing at this point) combine two elements into a single ionic compound.
For example, the reaction between aluminum and chlorine makes aluminum chloride.
Al + Cl2 → AlCl
Of course, we need to make sure that the formulas are written correctly. So, since aluminum has a charge of +3 and chloride has a charge of -1, the product will be AlCl3.
Al + Cl2 → AlCl3
Now, balancing will leave us with the finished reaction:
Al + 3 Cl2 → 2 AlCl3
In those cases where the metal has more than one possible charge
Cu + P4 → the metal will end up with the most common charge (+2 for copper). Given the -3 charge for phosphide, the reaction will be written
Cu + P4 → Cu3P2 and balanced
6 Cu + P4 → 2 Cu3P2
Decomposition Reactions Decomposition reactions are those in which a compound breaks down into simpler compounds or elements. Recognizing Decomposition Reactions Decomposition reactions are very easy to spot. They have one reactant. No other reactions have only one reactant. Of course, that doesn't mean the the coefficient has to be one, only that there is only one reactant.
2 NaCl → 2 Na + Cl2
2 KClO3 → 3 O2 + 2 KCl
Completing decomposition reactions Decomposition reactions fall into two general types, simple decompostion reactions and special cases. Simple decompositions are those that start with a simple ionic compound (containing only two elements). Special cases are several classes of compounds whose products can be predicted easily if you know that patterns.
Completing simple decomposition reactions To complete a simple decomposition reaction, simple break the compound into its elements (remembering the HOFBrINCl's) and then balance. For instance
KF → K + F turns into 2 KF → 2 K + F2 when you remember that F is a HOFBrINCl and then balance the reaction.
Other examples:
2 AlH3 → 2 Al + 3 H2
8 CaS → 8 Ca + S8
Completing special case decomposition reactions There are four common types of special decomposition reactions, so in order to correctly complete them, you must first recognize the type. They are: carbonates, such as K2CO3, MgCO3 and Al2(CO3)3, bicarbonates, such as KHCO3, Mg(HCO3)2 and Al(HCO3)2, chlorates, such as KClO3, Mg(ClO3)2 and Al(ClO3)3 and ammonium compounds, such as NH4Cl, (NH4)2SO4 and (NH4)3PO4
Carbonates Carbonates decompose to form carbon dioxide and an oxide. For instance: K2CO3 → CO2 + K2O MgCO3 → CO2 + MgO Al2(CO3)3 → 3 CO2 + Al2O3
Of course, you need to make sure that the charges work in the products. So we get K2O, not KO and Al2O3, not AlO.
Bicarbonates Bicarbonates decompose to form carbon dioxide, water and an oxide. For instance: 2 KHCO3 → H2O + 2 CO2 + K2O Mg(HCO3)2 → H2O + 2 CO2 + MgO 2 Al(HCO3)3 → 3 H2O + 6 CO2 + Al2O3
Chlorates Chlorates decompose to form oxygen gas (O2, remember...it's a HOFBrINCl) and a chloride. For instance: 2 KClO3 → 3 O2 + 2 KCl Mg(ClO3)2 → 3 O2 + MgCl2 2 Al(ClO3)3 → 9 O2 + 2 AlCl3
Ammonium compounds Ammonium compounds decompose to form ammonia (NH3) and an acid. For instance: NH4Cl → NH3 + HCl (NH4)2SO4 → 2 NH3 + H2SO4 (NH4)3PO4 → 3 NH3 + H3PO4