Balancing Double Displacement Reactions In a | StudyX - Free Printable

Let’s solve each single replacement reaction step by step.

In a single replacement reaction, one element replaces another in a compound. The general form is:

> A + BC → AC + B (if A is more reactive than B)

But if the replacing element is *less* reactive, no reaction occurs → we write “N.R.”

We’ll use the activity series logic:
- Metals higher in the activity series can replace metals lower down.
- For water or acids, very active metals (like Li, K, Ca, Na) react; less active ones don’t.
- We’re given charges for transition metals to help us write correct formulas.

Also note: HOH = H₂O (water), so reaction 5 is metal + water.

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1) Fe + CuCl₂ →

Iron (Fe²⁺) vs Copper (Cu⁺? Wait — problem says Copper: Cu⁺, but in CuCl₂, copper must be Cu²⁺ because Cl is -1 × 2 = -2 → so Cu is +2. But the worksheet says “for transition metals use the following charges” and lists Copper as Cu⁺. That seems inconsistent with CuCl₂.

Wait — let’s read carefully:
> “For transition metals use the following charges: Iron: Fe²⁺, Mercury: Hg²⁺, Lead: Pb⁴⁺, Copper: Cu⁺, Gold: Au³⁺”

But in CuCl₂, if Cl is -1, then two Cl⁻ means total -2, so Cu must be +2. So there’s a conflict.

However, the instruction says “use the following charges” — meaning when writing products, assign those charges to the ions. But that would make CuCl impossible? No — perhaps they mean: when the metal becomes an ion in product, use that charge.

Actually, looking at reaction 1: Fe + CuCl₂ → ?

Fe is trying to replace Cu in CuCl₂. In CuCl₂, Cu is +2 (because 2×Cl⁻ = -2). But the worksheet says “Copper: Cu⁺”. This is confusing.

Perhaps the charge list is for when the metal forms its own compound — i.e., when Fe becomes ion, it’s Fe²⁺; when Cu becomes ion, it’s Cu⁺ — but that contradicts CuCl₂.

I think there might be a typo in the worksheet. Because CuCl₂ implies Cu²⁺. But since the worksheet explicitly says “Copper: Cu⁺”, maybe they want us to assume copper always forms +1? Then CuCl₂ wouldn't exist — which doesn’t make sense.

Alternative interpretation: Maybe the charge list is only for when the metal is displaced and becomes a cation in the product — not for the original compound.

Let me check standard chemistry: In reality, Fe can displace Cu from CuCl₂ because Fe is above Cu in activity series. Product is FeCl₂ + Cu. And Fe is Fe²⁺, Cu is reduced to Cu⁰.

But according to the worksheet’s charge list, Copper is Cu⁺ — so maybe they expect Cu⁺ in products? But that would require different stoichiometry.

This is messy. Let’s look at other reactions.

Reaction 4: Pb + Au(NO₃)₃ → Gold is Au³⁺ per list. Nitrate is NO₃⁻, so Au(NO₃)₃ makes sense.

Reaction 6: K + AgCl → Silver isn’t listed, but typically Ag⁺.

I think the charge list is meant for when the metal appears as an ion in the *product*. So for example, if iron displaces something, it becomes Fe²⁺. If copper is displaced, it becomes Cu⁰ (elemental), so charge doesn’t matter. Only when it's in a compound.

In reaction 1: Fe + CuCl₂ → Fe will become Fe²⁺, so it needs two Cl⁻ → FeCl₂. Cu becomes elemental Cu.

So even though the worksheet says "Copper: Cu⁺", that might be irrelevant here because copper is being reduced to metal, not forming an ion.

Similarly, in reaction 9: Fe + Cu(OH)₂ → Fe displaces Cu, becomes Fe²⁺, so Fe(OH)₂, and Cu metal.

I think we should ignore the charge list for the displaced metal becoming elemental — it only matters when the displacing metal forms a compound.

Moreover, for consistency, let’s apply standard single replacement rules with common charges unless specified otherwise for the product ion.

But the worksheet says: “for transition metals use the following charges” — probably meaning when they form compounds in products.

So for Fe in product: Fe²⁺
For Cu in product: if it were to form a compound, it would be Cu⁺ — but in these reactions, Cu is usually displaced as metal, so no issue.

Except in reaction where Cu is the displacer? Like reaction 8: Cu + Fe(OH)₃ → here Cu is trying to displace Fe. But Cu is less reactive than Fe, so no reaction anyway.

Let’s proceed with standard activity series and use the given charges only when the metal ends up in a compound.

Standard activity series (most reactive to least):
Li, K, Ba, Ca, Na, Mg, Al, Zn, Fe, Ni, Sn, Pb, H, Cu, Ag, Au

So:

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1) Fe + CuCl₂ →

Fe is above Cu → reaction occurs.

Fe becomes Fe²⁺ (given), Cl is Cl⁻, so FeCl₂.

Cu becomes Cu(s).

Unbalanced: Fe + CuCl₂ → FeCl₂ + Cu

Already balanced! Atoms: Fe:1=1, Cu:1=1, Cl:2=2.

✔ Answer: Fe + CuCl₂ → FeCl₂ + Cu

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2) Hg + Sn(SO₄)₂ →

First, what is Sn(SO₄)₂? SO₄ is sulfate, charge -2. So two SO₄²⁻ = -4, so Sn must be +4. Tin(IV) sulfate.

Hg is mercury. Activity series: Hg is below Sn? Let’s see: Sn is above H in series, Hg is below H. So Sn is more reactive than Hg.

Therefore, Hg cannot displace Sn → no reaction.

✔ Answer: N.R.

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3) Ba + Ni₃(PO₄)₂ →

Barium is very reactive (above Ca). Nickel phosphate: PO₄ is -3, so Ni₃(PO₄)₂ means 3 Ni ions and 2 PO₄³⁻ → total negative charge: 2×(-3)= -6, so 3 Ni must be +6 total → each Ni is +2.

Ba is above Ni in activity series → reaction occurs.

Ba becomes Ba²⁺ (alkaline earth metal).

Product: Ba₃(PO)₂? Because PO₄ is -3, Ba is +2 → need 3 Ba²⁺ and 2 PO₄³⁻ to balance: 3×(+2)=+6, 2×(-3)=-6.

And Ni becomes Ni(s).

So unbalanced: Ba + Ni₃(PO₄)₂ → Ba₃(PO)₂ + Ni

Balance:

Left: Ba:1, Ni:3, P:2, O:8
Right: Ba:3, Ni:1, P:2, O:8

Need 3 Ba on left, 3 Ni on right.

So: 3Ba + Ni₃(PO₄)₂ → Ba₃(PO₄)₂ + 3Ni

Check: Ba:3=3, Ni:3=3, P:2=2, O:8=8 → balanced.

✔ Answer: 3Ba + Ni₃(PO₄)₂ → Ba₃(PO)₂ + 3Ni

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4) Pb + Au(NO₃)₃ →

Lead and gold nitrate.

Au(NO₃)₃: Au is +3 (given), NO₃ is -1, so 3×(-1)= -3, matches.

Activity series: Pb is above Au? Yes, Pb is before H, Au is last. So Pb can displace Au.

Pb becomes Pb⁴⁺? Worksheet says Lead: Pb⁴⁺. But typically lead is +2 in such reactions. However, worksheet instructs to use Pb⁴⁺.

So Pb⁴⁺ and NO₃⁻ → Pb(NO₃)₄? But that might not be stable, but per instructions, we use Pb⁴⁺.

Au becomes Au(s).

Unbalanced: Pb + Au(NO₃)₃ → Pb(NO₃)₄ + Au

But now atoms don’t match. Left: Pb:1, Au:1, N:3, O:9
Right: Pb:1, N:4, O:12, Au:1 → not balanced.

To balance, need same number of NO₃ groups.

Since Pb is +4, it needs 4 NO₃⁻. Au(NO₃)₃ has 3 NO₃⁻ per Au.

So find LCM of 3 and 4 is 12.

So 4 Au(NO₃)₃ provide 12 NO₃⁻ → requires 3 Pb to take 12 NO₃⁻ (since each Pb takes 4).

Then Au produced: 4 Au.

So: 3Pb + 4Au(NO₃)₃ → 3Pb(NO₃)₄ + 4Au

Check atoms:

Left: Pb:3, Au:4, N:12, O:36
Right: Pb:3, N:12 (from 3×4=12 NO₃), O:36, Au:4 → balanced.

But is this realistic? Probably not, but per worksheet instructions, we use Pb⁴⁺.

Alternatively, maybe the charge list is only for when the metal is in a compound as a result of displacement, but in this case, Pb is displacing, so it becomes ion.

I think we have to go with the instruction.

✔ Answer: 3Pb + 4Au(NO₃)₃ → 3Pb(NO₃)₄ + 4Au

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5) Li + HOH →

HOH is H₂O. Lithium reacts with water.

Very active metal → produces hydrogen gas and base.

General: 2M + 2H₂O → 2MOH + H₂

Li becomes Li⁺, OH⁻, so LiOH.

Unbalanced: Li + H₂O → LiOH + H₂

Balance: 2Li + 2H₂O → 2LiOH + H₂

Atoms: Li:2=2, H:4=4 (2 in 2LiOH is 2H, plus H₂ is 2H, total 4H), O:2=2.

✔ Answer: 2Li + 2H₂O → 2LiOH + H₂

(Note: HOH is same as H₂O)

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6) K + AgCl →

Potassium and silver chloride.

K is very reactive, Ag is not. K is above Ag in activity series → reaction occurs.

K becomes K⁺, Cl is Cl⁻, so KCl.

Ag becomes Ag(s).

Unbalanced: K + AgCl → KCl + Ag

Already balanced.

✔ Answer: K + AgCl → KCl + Ag

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7) Ca + NaOH →

Calcium and sodium hydroxide.

Can Ca displace Na? Activity series: Ca is above Na? Standard series: K, Na, Ca... wait no.

Standard activity series: Li, K, Ba, Ca, Na, ... so Ca is above Na? Actually, calcium is more reactive than sodium? No.

Recall: alkali metals are most reactive: Li > K > Ba > Sr > Ca > Na > Mg...

Actually, barium and calcium are alkaline earth, but still, sodium is less reactive than calcium? Let me confirm.

Standard reactivity series for metals:

Most reactive: Potassium (K), Sodium (Na), Calcium (Ca), Magnesium (Mg), Aluminum (Al), Zinc (Zn), Iron (Fe), Tin (Sn), Lead (Pb), Hydrogen (H), Copper (Cu), Silver (Ag), Gold (Au)

Wait, I think I have it wrong.

Correct order (decreasing reactivity):

Lithium (Li)
Potassium (K)
Barium (Ba)
Strontium (Sr)
Calcium (Ca)
Sodium (Na)
Magnesium (Mg)
Aluminum (Al)
Manganese (Mn)
Zinc (Zn)
Chromium (Cr)
Iron (Fe)
Cadmium (Cd)
Cobalt (Co)
Nickel (Ni)
Tin (Sn)
Lead (Pb)
Hydrogen (H)
Antimony (Sb)
Bismuth (Bi)
Copper (Cu)
Tungsten (W)
Mercury (Hg)
Silver (Ag)
Gold (Au)
Platinum (Pt)

So Calcium is above Sodium? In many sources, sodium is more reactive than calcium? No.

Actually, calcium reacts vigorously with water, sodium also, but sodium is more reactive? I'm confused.

Standard teaching: Group 1 metals (alkali) are more reactive than Group 2 (alkaline earth) in the same period, but calcium is below sodium in group? No, sodium is group 1, calcium group 2.

Reactivity increases down a group and decreases across a period.

Sodium (group 1, period 3) vs Calcium (group 2, period 4) — calcium is in higher period, so should be more reactive? But actually, alkali metals are more reactive than alkaline earth in adjacent periods.

Upon recall: Potassium > Sodium > Lithium > Barium > Strontium > Calcium > Magnesium...

I think sodium is more reactive than calcium.

Confirm with water reaction: sodium reacts violently with cold water, calcium reacts moderately with cold water, magnesium with hot water.

So sodium is more reactive than calcium.

Therefore, calcium cannot displace sodium from NaOH.

Because Ca is less reactive than Na.

So no reaction.

Is that correct? Let me double-check.

In some sources, the activity series is: K, Na, Ca, Mg, Al, Zn, Fe, Sn, Pb, H, Cu, Ag, Au

Here Ca is after Na, meaning Na is more reactive.

Yes, typically sodium is placed above calcium.

For example, sodium displaces hydrogen from water easily, calcium does too but slower? Actually both do, but sodium is faster.

But in terms of displacement: can Ca displace Na from compound? Since Na is more reactive, no.

The rule is: a metal can displace another metal from solution only if it is higher in the activity series.

If Na is above Ca, then Ca cannot displace Na.

In standard series, sodium is more reactive than calcium.

For instance, in electrolysis, sodium is harder to reduce than calcium? No.

Standard reduction potentials:

Na⁺/Na: -2.71 V
Ca²⁺/Ca: -2.87 V

More negative means more reactive (easier to oxidize).

Ca²⁺ + 2e⁻ → Ca E° = -2.87 V
Na⁺ + e⁻ → Na E° = -2.71 V

So calcium has more negative reduction potential, meaning it is more easily oxidized, so more reactive than sodium.

Oh! I had it backward.

Reduction potential more negative → stronger reducing agent → more reactive metal.

So Ca (-2.87 V) is more reactive than Na (-2.71 V).

Therefore, calcium can displace sodium from compounds.

Is that true? In practice, calcium does react with sodium compounds? I'm not sure.

But theoretically, based on reduction potentials, yes.

In many textbooks, the activity series lists calcium above sodium.

For example: Li, K, Ba, Ca, Na, Mg, ...

Yes, barium and calcium are above sodium.

So Ca is more reactive than Na.

Therefore, Ca can displace Na from NaOH.

But NaOH is a base, and calcium would react with water part.

NaOH is aqueous, I assume, so it's Na⁺ and OH⁻ in water.

Calcium will react with water first, producing Ca(OH)₂ and H₂, and then possibly with NaOH, but since NaOH is already there, it might not change much.

In single replacement, if we consider Ca + NaOH, it could be seen as Ca displacing Na, but NaOH contains Na⁺ and OH, so product would be Ca(OH)₂ and Na.

But Ca(OH)₂ is slightly soluble, and Na is metal.

However, in aqueous solution, sodium metal would immediately react with water, so it's complicated.

But for the purpose of this worksheet, we treat it as a direct displacement.

Since Ca is more reactive than Na, reaction should occur.

Ca becomes Ca²⁺, OH⁻ is OH⁻, so Ca(OH)₂.

Na becomes Na(s).

Unbalanced: Ca + NaOH → Ca(OH)₂ + Na

Balance: need 2 NaOH to provide 2 OH⁻ for Ca(OH)₂.

So: Ca + 2NaOH → Ca(OH)₂ + 2Na

Atoms: Ca:1=1, Na:2=2, O:2=2, H:2=2 → balanced.

But is this correct? In reality, sodium metal produced would react with water, but perhaps for this level, we write it as is.

Some might argue that since NaOH is in water, Ca reacts with water instead, but the compound is NaOH, so we consider the ion.

I think for consistency, since Ca is more reactive, we say reaction occurs.

Moreover, in the activity series, Ca is above Na.

So ✔ Answer: Ca + 2NaOH → Ca(OH)₂ + 2Na

But let's confirm with standard knowledge. Upon second thought, in aqueous solution, calcium will react with water to give Ca(OH)2 and H2, and the Na+ remains, so no displacement of sodium metal. But the worksheet might expect the theoretical displacement.

Looking back at reaction 5: Li + HOH → which is water, and we wrote LiOH and H2, not displacing anything else.

Here, NaOH is not water; it's a compound containing Na.

But in solution, it's dissociated.

Perhaps for single replacement with bases, it's similar to water if the metal is very active.

But the problem is "single replacement reactions", and typically for metals with bases, if the metal is above hydrogen, it may react, but displacing the cation.

I recall that very active metals like K, Na, Ca react with water, not necessarily displacing other metals from their hydroxides.

For example, calcium added to sodium hydroxide solution will react with water to produce calcium hydroxide and hydrogen gas, and sodium ions remain.

So no displacement of sodium metal.

Therefore, perhaps no reaction in the sense of displacing Na, but there is a reaction with water.

The worksheet likely intends for us to consider whether the metal can displace the cation in the compound.

Given that, and since Ca is more reactive than Na, theoretically it should, but practically it doesn't because of water.

To resolve, let's look at the answer expected.

In many worksheets, for Ca + NaOH, they say no reaction because sodium is more reactive? But we saw reduction potential shows Ca is more reactive.

Perhaps the activity series used in schools places Na above Ca.

Upon checking my memory, in many high school texts, the series is: K, Na, Ca, Mg, Al, Zn, Fe, Sn, Pb, H, Cu, Ag, Au — so Na before Ca, meaning Na is more reactive.

For example, sodium reacts explosively with water, calcium reacts steadily, so sodium is considered more reactive.

Reduction potential: Na -2.71, Ca -2.87, so Ca is more reactive, but perhaps in terms of kinetics or something, sodium is faster.

I think for this context, since the worksheet includes Ba in reaction 3, and Ba is very reactive, and for Ca + NaOH, if we follow the series where Na is above Ca, then no reaction.

Let's assume the standard school activity series: Li, K, Ba, Ca, Na, Mg, ... no, usually it's K, Na, Li, Ba, Ca, etc.

I found a reliable source in my mind: the activity series is often listed as:

Potassium (K)
Sodium (Na)
Lithium (Li)
Barium (Ba)
Strontium (Sr)
Calcium (Ca)
Magnesium (Mg)
...

So sodium is above calcium.

Therefore, calcium cannot displace sodium.

So for Ca + NaOH, no reaction.

Moreover, in the compound NaOH, the cation is Na+, and Ca is less reactive than Na, so no displacement.

So ✔ Answer: N.R.

I'll go with that to match common textbook approach.

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8) Cu + Fe(OH)₃ →

Copper and iron(III) hydroxide.

Activity series: Cu is below Fe. Fe is above H, Cu is below H, so Fe is more reactive than Cu.

Therefore, Cu cannot displace Fe.

No reaction.

✔ Answer: N.R.

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9) Fe + Cu(OH)₂ →

Iron and copper(II) hydroxide.

Fe is above Cu in activity series → reaction occurs.

Fe becomes Fe²⁺ (given), OH⁻ is OH⁻, so Fe(OH)₂.

Cu becomes Cu(s).

Unbalanced: Fe + Cu(OH)₂ → Fe(OH)₂ + Cu

Already balanced.

Atoms: Fe:1=1, Cu:1=1, O:2=2, H:2=2.

✔ Answer: Fe + Cu(OH)₂ → Fe(OH)₂ + Cu

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Now, compiling all answers:

1) Fe + CuCl₂ → FeCl₂ + Cu
2) Hg + Sn(SO₄)₂ → N.R.
3) 3Ba + Ni₃(PO₄)₂ → Ba₃(PO₄)₂ + 3Ni
4) 3Pb + 4Au(NO₃)₃ → 3Pb(NO₃)₄ + 4Au
5) 2Li + 2H₂O → 2LiOH + H₂ (since HOH = H₂O)
6) K + AgCl → KCl + Ag
7) Ca + NaOH → N.R. (assuming Na is more reactive)
8) Cu + Fe(OH)₃ → N.R.
9) Fe + Cu(OH)₂ → Fe(OH)₂ + Cu

For reaction 5, the worksheet writes HOH, so perhaps write HOH in answer, but it's the same as H₂O. Usually we write H₂O, but to match, maybe keep HOH.

In the problem, it's written as HOH, so in answer, we can write HOH.

Similarly, in balancing, 2Li + 2HOH → 2LiOH + H₂

Yes.

For reaction 4, using Pb⁴⁺ as instructed.

Final Answer:

1) Fe + CuCl₂ → FeCl₂ + Cu
2) N.R.
3) 3Ba + Ni₃(PO₄)₂ → Ba₃(PO₄)₂ + 3Ni
4) 3Pb + 4Au(NO₃)₃ → 3Pb(NO₃)₄ + 4Au
5) 2Li + 2HOH → 2LiOH + H₂
6) K + AgCl → KCl + Ag
7) N.R.
8) N.R.
9) Fe + Cu(OH)₂ → Fe(OH)₂ + Cu