Types of reactions — combination, decomposition, displacement, double displacement, redox: Exercises

Before we start, one shared picture — the decision path every question below can be walked through:

Figure — Types of reactions — combination, decomposition, displacement, double displacement, redox

Read it like a flowchart: count reactants and products first (how many boxes go in, how many come out). That single count already splits combination from decomposition. Then ask "is there a lone element?" to spot displacement, and "did any oxidation number move?" to overlay redox.

LEVEL 1 — Recognition

Goal: given a balanced equation, name the pattern. No prediction yet.

Recall Solution Q1

WHAT to notice: two separate substances ( $H_{2}$ and $O_{2}$ ) go in, one substance ( $H_{2} O$ ) comes out. WHY that answer: "two or more → one product" is the skeleton $A + B \to A B$ . Answer: Combination reaction. (It is also redox — H goes $0 \to + 1$ , O goes $0 \to - 2$ — but at L1 we only name the shape.)

Recall Solution Q2

WHAT to notice: two compounds in, two compounds out, and the partners have swapped — $A g$ left the $N O_{3}$ and joined $C l$ ; $N a$ did the reverse. WHY: skeleton $A B + C D \to A D + C B$ . Answer: Double displacement (a precipitation, since $A g C l$ is an insoluble solid).

Recall Solution Q3

WHAT to notice: exactly one compound goes in, two come out, and heat ( $Δ$ ) is supplied. WHY: "one → two or more, needs energy" is the decomposition skeleton $A B \to A + B$ ; the heat symbol names the sub-type. Answer: Thermal decomposition (thermolysis).

LEVEL 2 — Application

Goal: predict the products AND balance the equation yourself.

Recall Solution Q4

WHAT: A lone metal ( $Z n$ ) meets a compound ( $H C l$ ) ⇒ displacement, $A + B C \to A C + B$ . WHY it goes: In the Reactivity Series of Metals, $Z n$ sits above hydrogen, so $Z n$ displaces $H$ . The metal takes the chloride; hydrogen leaves as gas. Skeleton products: $Z n C l_{2} + H_{2}$ . Balance (Balancing Chemical Equations): $Z n$ needs two $C l$ , so put 2 on $H C l$ ; that gives 2 H ⇒ one $H_{2}$ . $Z n + 2 H C l \to Z n C l_{2} + H_{2} ↑$ Check: Zn 1=1, H 2=2, Cl 2=2. ✔

Recall Solution Q5

WHAT: lone metal + salt ⇒ displacement. WHY: $A l$ is above $C u$ in the reactivity series, so $A l$ kicks $C u$ out. The subtlety — charges differ: $A l$ forms $A l^{3 +}$ but $C u$ was $C u^{2 +}$ . The sulfate is $S O_{4}^{2 -}$ . So aluminium sulfate is $A l_{2} (S O_{4})_{3}$ (two $A l^{3 +}$ = $+ 6$ balances three $S O_{4}^{2 -}$ = $- 6$ ). Balance: need three $S O_{4}$ ⇒ 3 $C u S O_{4}$ and 3 $C u$ ; that needs 2 $A l$ . $2 A l + 3 C u S O_{4} \to A l_{2} (S O_{4})_{3} + 3 C u$ Check: Al 2=2, Cu 3=3, S 3=3, O 12=12. ✔

Recall Solution Q6

WHAT: one compound, heated ⇒ thermal decomposition. Metal nitrates of moderately reactive metals give metal oxide + nitrogen dioxide + oxygen. Skeleton products: $P b O + N O_{2} + O_{2}$ . Balance: Start with one $P b$ . Two $N O_{3}$ give 2 N ⇒ $2 N O_{2}$ . Count O: left has $2 \times 3 = 6$ ; right has $P b O$ (1) $+ 2 N O_{2}$ (4) $= 5$ , leaving 1 O ⇒ that is $\frac{1}{2} O_{2}$ . Clear the fraction by doubling everything: $2 P b (N O_{3})_{2} Δ 2 P b O + 4 N O_{2} ↑ + O_{2} ↑$ Check: Pb 2=2, N 4=4, O 12 = (2 + 8 + 2) = 12. ✔

LEVEL 3 — Analysis

Goal: assign oxidation numbers and decide whether the reaction is redox. Reference: Oxidation Number Rules.

Recall Solution Q7

Assign (rule: free element = 0; ion = its charge):

$F e$ (free metal) $= 0$ → in $F e S O_{4}$ , iron is $F e^{2 +}$ $= + 2$ .
$C u$ in $C u S O_{4}$ is $C u^{2 +}$ $= + 2$ → free $C u = 0$ .
$S O_{4}^{2 -}$ unchanged on both sides (spectator).

WHAT moved: Fe: $0 \to + 2$ (number rose ⇒ lost electrons ⇒ oxidised). Cu: $+ 2 \to 0$ (number fell ⇒ gained electrons ⇒ reduced). Use OIL RIG. Redox? Electron numbers changed ⇒ YES, redox. Agents: the one reduced is the oxidising agent ⇒ $C u^{2 +}$ (in $C u S O_{4}$ ). The one oxidised is the reducing agent ⇒ $F e$ . Electron bookkeeping: Fe loses 2 e⁻, Cu²⁺ gains 2 e⁻ — equal, so charge is conserved.

Recall Solution Q8

Assign every element on both sides:

$B a$ : $+ 2$ (in $B a C l_{2}$ ) → $+ 2$ (in $B a S O_{4}$ ). No change.
$C l$ : $- 1$ → $- 1$ (in $N a C l$ ). No change.
$N a$ : $+ 1$ → $+ 1$ . No change.
$S$ in $S O_{4}$ : $+ 6$ → $+ 6$ . No change. (O stays $- 2$ throughout.)

WHAT this means: not a single oxidation number moved. Redox? NO. No electron transfer — the ions merely changed partners. Lesson: double displacements are almost never redox; this confirms the parent's warning.

Recall Solution Q9

Assign in $H_{2} O$ : H $= + 1$ , O $= - 2$ . In the free gases both are $0$ .

H: $+ 1 \to 0$ ⇒ gained electrons ⇒ reduced ⇒ produced at the cathode (negative electrode attracts $H^{+}$ ).
O: $- 2 \to 0$ ⇒ lost electrons ⇒ oxidised ⇒ produced at the anode.

Redox? Yes. Cathode product = $H_{2}$ . Volume ratio $H_{2} : O_{2} = 2 : 1$ .

LEVEL 4 — Synthesis

Goal: combine pattern-recognition, balancing, and the redox lens in one problem.

Recall Solution Q10

(a) Displacement (Fe above Cu): $F e + C u S O_{4} \to F e S O_{4} + C u$ Already balanced (Fe 1=1, Cu 1=1, S 1=1, O 4=4). (b) Shape axis: displacement. Redox axis: yes (Fe $0 \to + 2$ , Cu $+ 2 \to 0$ ). So it is a displacement that is also redox. (c) Blue $C u S O_{4}$ fades toward pale green $F e S O_{4}$ ; a reddish-brown copper coating forms on the nail. (d) $C u + F e S O_{4} \to$ no reaction, because Cu is below Fe in the Reactivity Series of Metals — a less reactive metal cannot displace a more reactive one. The reverse failing confirms the forward direction.

Recall Solution Q11

(a) Combination and redox: $2 M g + O_{2} \to 2 M g O$ (b) Moles of Mg: $\frac{12.0}{24} = 0.5 mol$ . Ratio $M g : M g O = 2 : 2 = 1 : 1$ ⇒ $0.5 mol M g O$ . Mass of MgO $= 0.5 \times 40 = 20.0 g$ . Ratio $M g : O_{2} = 2 : 1$ ⇒ $O_{2} = 0.25 mol$ . Mass of $O_{2}$ $= 0.25 \times 32 = 8.0 g$ . Conservation check: $12.0 g M g + 8.0 g O_{2} = 20.0 g M g O$ . Mass in = mass out. ✔

LEVEL 5 — Mastery

Goal: design, justify, and generalise — reasoning beyond a single equation.

Recall Solution Q12

Test 1 — add dilute $H C l$ : only the carbonate fizzes (double displacement producing a gas): $N a_{2} C O_{3} + 2 H C l \to 2 N a C l + H_{2} O + C O_{2} ↑$ (The unstable $H_{2} C O_{3}$ decomposes to $H_{2} O + C O_{2}$ .) ⇒ the one that fizzes is $N a_{2} C O_{3}$ . Test 2 — add $B a C l_{2}$ to the remaining two: only the sulfate gives a white precipitate (double displacement, precipitation): $N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} ↓ + 2 N a C l$ ⇒ that one is $N a_{2} S O_{4}$ . By elimination: the last, which does neither, is $N a C l$ . Redox check: neither reaction changes any oxidation number ⇒ both are non-redox double displacements — expected for ion-swap tests.

Recall Solution Q13

Slaking of lime: assign numbers — Ca stays $+ 2$ , O stays $- 2$ , H stays $+ 1$ across the arrow. No number moves ⇒ not redox, even though two things became one. It is exothermic (Exothermic and Endothermic Reactions) but heat release does not imply redox. Burning Mg: Mg $0 \to + 2$ , O $0 \to - 2$ ⇒ numbers move ⇒ redox. Generalisation: A combination is also redox exactly when at least one reactant is a free element (oxidation number 0) that ends up combined. A free element must change its oxidation number to bond, forcing electron transfer. When both reactants are already compounds with fixed ionic charges (like $C a O$ and $H_{2} O$ ), no free element exists to change, so no redox.

Recall Solution Q14

Read each clue as an inequality:

Displaces H from acid ⇒ $M$ is above H.
Displaces Cu ⇒ $M$ is above Cu (already implied, since Cu is below H).
Does not displace Mg ⇒ $M$ is below Mg.

Combine: $M g > M > H$ . So $M$ lies in the band ${A l, Z n, F e, P b}$ — below Mg and above hydrogen. (We cannot pin one element without more data, but the range is fixed.) A valid divalent equation (take $M$ acting on copper sulfate): $M + C u S O_{4} \to M S O_{4} + C u$ Balanced 1:1:1:1 because $M^{2 +}$ matches $C u^{2 +}$ . This is displacement and redox ( $M : 0 \to + 2$ , $C u : + 2 \to 0$ ).

Wrap-up recall

Recall One-line pattern reminders

Displacement direction rule ::: The lone element must be more reactive than the one it displaces; failure means it is less reactive. Test for redox ::: Assign oxidation numbers on both sides — redox iff at least one number changes. Gram-to-mole rule in stoichiometry ::: Coefficients are mole ratios; convert grams to moles before applying them. When a combination is also redox ::: When a free element (ox. no. 0) is a reactant. What drives a double displacement ::: A product leaves solution — precipitate, gas, or water.