Reaction Types Practice

Two skills, tangled together. Balancing is arithmetic and you can check it yourself in ten seconds by counting. Classifying is a judgement about the shape of the equation, and the shape is what lets you predict products for a reaction you have never seen.

Balance by changing coefficients only. Changing a subscript changes the substance, which is not balancing — it is inventing.

1. Balance each equation and name its type. (a) $Mg + O X_{2} MgO$ (b) $H X_{2} O X_{2} H X_{2} O + O X_{2}$ (c) $Zn + HCl ZnCl X_{2} + H X_{2}$ (d) $AgNO X_{3} + NaCl AgCl + NaNO X_{3}$

Answer 1

(a) $2 Mg + O X_{2} 2 MgO$

Oxygen arrives as a diatomic molecule, so there are two oxygen atoms on the left and the right side must have an even number too. Two magnesium oxides need two magnesiums. Count to check: Mg 2 and 2, O 2 and 2. Synthesis — two substances in, one out.

(b) $2 H X_{2} O X_{2} 2 H X_{2} O + O X_{2}$

Count: H 4 and 4, O 4 and $2 + 2 = 4$ . Decomposition — one substance in, more than one out. This is the reaction you ran at station B in Sorting Five Reactions.

(c) $Zn + 2 HCl ZnCl X_{2} + H X_{2}$

Count: Zn 1 and 1, H 2 and 2, Cl 2 and 2. Single displacement — an element and a compound in, a different element and a different compound out. Zinc has taken chlorine’s partner away from hydrogen.

(d) Already balanced as written: $AgNO X_{3} + NaCl AgCl + NaNO X_{3}$

Count: Ag 1 and 1, Na 1 and 1, Cl 1 and 1, and the nitrate travels intact so you can count it as one unit on each side. Double displacement — two compounds swap partners.

Counting polyatomic ions as units, rather than atom by atom, is the single biggest time-saver in balancing. It only works when the ion survives the reaction intact, which in a double displacement it usually does.

2. Balance these combustion equations. (a) $C X_{3} H X_{8} + O X_{2} CO X_{2} + H X_{2} O$ (b) $C X_{2} H X_{6} + O X_{2} CO X_{2} + H X_{2} O$ (c) $C X_{2} H X_{5} OH + O X_{2} CO X_{2} + H X_{2} O$

Answer 2

The reliable order for a hydrocarbon: carbon first, hydrogen second, oxygen last, because oxygen appears in two products and is the one you want free to adjust at the end.

(a) Three carbons, so 3 $CO X_{2}$ . Eight hydrogens, so 4 $H X_{2} O$ . Now count the oxygen on the right: $(3 \times 2) + (4 \times 1) = 10$ , which needs 5 $O X_{2}$ .

$C X_{3} H X_{8} + 5 O X_{2} 3 CO X_{2} + 4 H X_{2} O$

(b) Two carbons, so 2 $CO X_{2}$ . Six hydrogens, so 3 $H X_{2} O$ . Oxygen on the right: $4 + 3 = 7$ , which needs 3.5 $O X_{2}$ . Fractions are legitimate chemistry and untidy convention, so double everything:

$2 C X_{2} H X_{6} + 7 O X_{2} 4 CO X_{2} + 6 H X_{2} O$

Check: C 4 and 4, H 12 and 12, O 14 and $8 + 6 = 14$ .

(c) Ethanol has an oxygen of its own, which is the whole point of this one. Two carbons give 2 $CO X_{2}$ ; six hydrogens give 3 $H X_{2} O$ . Right-hand oxygen: $4 + 3 = 7$ . But one of those seven came in with the ethanol, so only six have to be supplied by $O X_{2}$ :

$C X_{2} H X_{5} OH + 3 O X_{2} 2 CO X_{2} + 3 H X_{2} O$

Check: C 2 and 2, H 6 and 6, O $1 + 6 = 7$ and $4 + 3 = 7$ .

Forgetting the oxygen already inside the fuel is the standard error here, and it always produces a coefficient that is too large by exactly a half.

3. Predict the products, then balance. (a) $CaO + H X_{2} O \to$ (b) $SO X_{3} + H X_{2} O \to$ (c) $CaCO X_{3} \to$ (strongly heated) (d) $H X_{2} O \to$ (electrical decomposition)

Answer 3

(a) A metal oxide plus water gives the metal hydroxide.

$CaO + H X_{2} O Ca (OH) X_{2}$

Already balanced, and the solution is basic — which is what you saw with magnesium oxide in Oxides and Neutralisation.

(b) A non-metal oxide plus water gives an acid. Sulfur is in its higher combining state here, so the acid is sulfuric:

$SO X_{3} + H X_{2} O H X_{2} SO X_{4}$

Balanced as written: H 2 and 2, S 1 and 1, O $3 + 1 = 4$ and 4.

(c) A metal carbonate decomposes on strong heating to the metal oxide and carbon dioxide:

$CaCO X_{3} CaO + CO X_{2}$

This is the reaction in a lime kiln and it is one of the largest deliberate sources of carbon dioxide in industry.

(d) A binary compound decomposes into its elements, remembering that both are diatomic:

$2 H X_{2} O 2 H X_{2} + O X_{2}$

Check: H 4 and 4, O 2 and 2. Note that the products come out in a volume ratio of two to one, which is a result you can see directly in an electrolysis tube and which is not a coincidence.

4. For each pair, say whether a reaction occurs. If it does, write the balanced equation; if it does not, say what rules it out. (a) $Zn + CuSO X_{4}$ (b) $Cu + ZnSO X_{4}$ (c) $Mg + HCl$ (d) $Cu + HCl$ (e) $Cl X_{2} + NaBr$

Answer 4

Single displacement is decided by position in The Activity Series: a metal displaces one below it and cannot displace one above it.

(a) Yes. Zinc is above copper, so it takes the sulfate. $Zn + CuSO X_{4} ZnSO X_{4} + Cu$ This is the darkening strip you saw in Building the Activity Series.

(b) No reaction. Copper is below zinc, so it cannot displace it. The two would simply sit in the same beaker.

(c) Yes. Magnesium is above hydrogen, so it displaces hydrogen from the acid. $Mg + 2 HCl MgCl X_{2} + H X_{2}$

(d) No reaction. Copper sits below hydrogen in the series, so it will not displace hydrogen from a dilute acid. This is why copper plumbing survives contact with mildly acidic water and why the copper well in your spot plate stayed clear.

(e) Yes, and it uses the halogen series rather than the metal one. A more reactive halogen displaces a less reactive one, and chlorine outranks bromine. $Cl X_{2} + 2 NaBr 2 NaCl + Br X_{2}$ This is the demonstration you watched in the fume hood, and the orange colour that appeared was the bromine being set free.

5. Balance these and name the type. (a) $Na X_{2} CO X_{3} + CaCl X_{2} CaCO X_{3} + NaCl$ (b) $H X_{2} SO X_{4} + NaOH Na X_{2} SO X_{4} + H X_{2} O$

Answer 5

(a) $Na X_{2} CO X_{3} + CaCl X_{2} CaCO X_{3} + 2 NaCl$

Count: Na 2 and 2, C 1 and 1, O 3 and 3, Ca 1 and 1, Cl 2 and 2. Double displacement, and specifically a precipitation — calcium carbonate is insoluble, which is why it leaves the solution as a solid and why the reaction goes anywhere at all. This is the reaction behind Percentage Yield of a Precipitate.

(b) $H X_{2} SO X_{4} + 2 NaOH Na X_{2} SO X_{4} + 2 H X_{2} O$

Count: H $2 + 2 = 4$ and 4, S 1 and 1, Na 2 and 2, O $4 + 2 = 6$ and $4 + 2 = 6$ . Double displacement, and specifically a neutralisation — acid plus base gives a salt and water.

The coefficient 2 on the sodium hydroxide is not a balancing trick. Sulfuric acid supplies two acidic hydrogens per molecule, so it takes twice as much base to neutralise it as it would for hydrochloric acid at the same concentration. That fact will decide the arithmetic of a titration question later.

6. Hydrogen burning in oxygen gives water. (a) Write the balanced equation. (b) This reaction fits two of the five categories at once. Name them, and say whether that is a problem for the classification scheme. (c) Now classify $2 Mg + CO X_{2} 2 MgO + C$ and defend your answer.

Answer 6

(a) $2 H X_{2} + O X_{2} 2 H X_{2} O$

(b) It is a synthesis — two substances in, one out — and it is also a combustion, because a substance is reacting rapidly with oxygen and releasing energy.

That overlap is not a flaw in the scheme, because the scheme was never a set of boxes with no shared walls. Four of the categories describe the shape of the equation: how many substances go in, how many come out, what swaps with what. Combustion describes something different — what the reaction is doing chemically, namely burning in oxygen. A classification by shape and a classification by chemistry are answering different questions, so of course some reactions satisfy both.

The useful habit is to ask which categorisation earns its keep for the question in front of you. If you are predicting products, the shape categories do the work. If you are thinking about energy, safety, or what comes out of a chimney, “combustion” is the word that matters.

(c) Magnesium burning in carbon dioxide is a single displacement: an element (Mg) and a compound ( $CO X_{2}$ ) go in, and a different element (C) and a different compound (MgO) come out. Magnesium has displaced carbon from its oxide.

It is worth sitting with how strange that is. Carbon dioxide is what we use to put fires out, and magnesium burns in it — which is exactly why a carbon dioxide extinguisher is the wrong tool for a burning metal. A category tells you the pattern; it does not tell you what is safe.

7. Three claims from a class discussion. Correct each one, and say what the student was probably thinking. (a) “I balanced $Mg + O X_{2} MgO$ by writing $Mg + O X_{2} MgO X_{2}$ .” (b) ” $Fe + CuSO X_{4} FeSO X_{4} + Cu$ is a double displacement, because two things swapped.” (c) ” $NaCl + KNO X_{3} NaNO X_{3} + KCl$ — I mixed them and the equation balances, so a double displacement happened.”

Answer 7

(a) The equation now balances and describes a different substance. $MgO X_{2}$ is not magnesium oxide; magnesium is $Mg X^{2 +}$ and oxide is $O X^{2 -}$ , so the compound that exists is $MgO$ . Balancing is done with coefficients, which say how many, never with subscripts, which say what. The correct answer is $2 Mg + O X_{2} 2 MgO$ .

The student was doing something reasonable — looking for the fastest route to equal counts — and the fastest route is not always a legal one. This is the most common single error in the topic.

(b) It is a single displacement. The test is not “did things swap”; it is what kinds of things went in. Here an element (Fe) and a compound ( $CuSO X_{4}$ ) went in, and an element and a compound came out. A double displacement needs two compounds in and two compounds out, with no free element anywhere. Iron on the left and copper on the right are both elements, and that settles it.

(c) Nothing happened. The student is right that the equation balances, and that is not the question. All four of those substances are soluble, so what is actually in the beaker is a mixture of sodium, potassium, chloride, and nitrate ions, before and after, in the same proportions. Nothing left the solution as a solid, nothing left as a gas, and no water was formed — so there is no product to drive the reaction and no change to observe.

The general point is worth more than the example. A balanced equation is not evidence that a reaction occurs. You can write a balanced equation for almost anything. Whether it happens is decided by chemistry — solubility, the activity series, whether a gas or water is formed — and never by whether the arithmetic works out.

Reference: Types of Chemical Reactions and Predicting Products. Where you sorted these families yourself, before anyone named them: Sorting Five Reactions.

Curriculum connection

C2.1
use appropriate terminology related to chemical reactions, including, but not limited to: neutralization, precipitate, acidic, and basic [C]
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C2.2
write balanced chemical equations to represent synthesis, decomposition, single displacement, double displacement, and combustion reactions, using the IUPAC nomenclature system [PR, AI, C]
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Reaction Types Practice

Curriculum connection

C2.1

C2.2

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