49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable

Let’s go through each equation one by one and balance them. We’ll make sure the number of atoms of each element is the same on both sides.

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1) NaCl + KOH → NaOH + KCl

Left: Na=1, Cl=1, K=1, O=1, H=1
Right: Na=1, OH=1 (so O=1, H=1), K=1, Cl=1
→ Already balanced!

✔ Answer: 1, 1, 1, 1

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2) MgO + Li₂SO₄ → MgSO₄ + Li₂O

Left: Mg=1, O=1 from MgO + O=4 from SO₄ = total O=5? Wait — better to count per compound.

Actually:
- Left: Mg=1, O=1 (from MgO) + Li=2, S=1, O=4 (from Li₂SO₄) → Total: Mg=1, Li=2, S=1, O=5
- Right: Mg=1, S=1, O=4 (from MgSO₄) + Li=2, O=1 (from Li₂O) → Total: Mg=1, Li=2, S=1, O=5

→ Already balanced!

✔ Answer: 1, 1, 1, 1

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3) H₂O → H₂ + O₂

Left: H=2, O=1
Right: H=2, O=2 → Oxygen not balanced.

We need 2 H₂O to get 2 oxygen atoms on left → then we get 4 H atoms → so we need 2 H₂ on right.

Try:
2 H₂O → 2 H₂ + O₂
Left: H=4, O=2
Right: H=4, O=2 → Balanced!

✔ Answer: 2, 2, 1

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4) RbF + Be(NO₃)₂ → RbNO₃ + BeF₂

Left: Rb=1, F=1, Be=1, N=2, O=6
Right: Rb=1, NO₃=1 → N=1, O=3; Be=1, F=2 → F mismatch.

We need 2 RbF to give 2 F for BeF₂ → then we get 2 Rb → so we need 2 RbNO₃.

Try:
2 RbF + Be(NO₃)₂ → 2 RbNO₃ + BeF₂
Left: Rb=2, F=2, Be=1, N=2, O=6
Right: Rb=2, N=2, O=6, Be=1, F=2 → Balanced!

✔ Answer: 2, 1, 2, 1

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5) Ag + Cu(NO₃)₂ → AgNO₃ + Cu

Left: Ag=1, Cu=1, N=2, O=6
Right: Ag=1, NO₃=1 → N=1, O=3; Cu=1 → Nitrogen and oxygen not balanced.

We need 2 AgNO₃ to match 2 NO₃ groups → so we need 2 Ag on left.

Try:
2 Ag + Cu(NO₃)₂ → 2 AgNO₃ + Cu
Left: Ag=2, Cu=1, N=2, O=6
Right: Ag=2, N=2, O=6, Cu=1 → Balanced!

✔ Answer: 2, 1, 2, 1

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6) CO₂ + Cl₂ → CCl₄ + O₂

Left: C=1, O=2, Cl=2
Right: C=1, Cl=4, O=2 → Chlorine not balanced.

Need 2 Cl₂ to get 4 Cl atoms → then left has Cl=4.

But now:
CO₂ + 2 Cl₂ → CCl₄ + O₂
Left: C=1, O=2, Cl=4
Right: C=1, Cl=4, O=2 → Balanced!

Wait — is that right? Yes.

✔ Answer: 1, 2, 1, 1

*(Note: This reaction isn’t realistic chemically, but for balancing purposes, it works.)*

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7) CuSO₄ + HCN → Cu(CN)₂ + H₂SO₄

Left: Cu=1, S=1, O=4, H=1, C=1, N=1
Right: Cu=1, CN=2 → C=2, N=2; H=2, S=1, O=4 → Hydrogen, carbon, nitrogen not balanced.

We need 2 HCN to provide 2 H, 2 C, 2 N.

Try:
CuSO₄ + 2 HCN → Cu(CN)₂ + H₂SO₄
Left: Cu=1, S=1, O=4, H=2, C=2, N=2
Right: Cu=1, C=2, N=2, H=2, S=1, O=4 → Balanced!

✔ Answer: 1, 2, 1, 1

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8) Ga₂O₃ + Li → Li₂O + Ga

Left: Ga=2, O=3, Li=1
Right: Li=2, O=1, Ga=1 → All unbalanced.

Start with Ga: need 2 Ga on right → so put 2 in front of Ga.

Now: Ga₂O₃ + Li → Li₂O + 2 Ga

Oxygen: left=3, right=1 → need 3 Li₂O to get 3 O → then Li becomes 6 on right.

So: Ga₂O₃ + ? Li → 3 Li₂O + 2 Ga

Li on right = 6 → so need 6 Li on left.

Check:
Ga₂O₃ + 6 Li → 3 Li₂O + 2 Ga
Left: Ga=2, O=3, Li=6
Right: Li=6, O=3, Ga=2 → Balanced!

✔ Answer: 1, 6, 3, 2

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9) BaF₂ + Na₂S → BaS₂ + NaF

Wait — BaS₂? That’s unusual. Usually barium sulfide is BaS. But let’s follow the problem as written.

Left: Ba=1, F=2, Na=2, S=1
Right: Ba=1, S=2, Na=1, F=1 → Everything off.

If product is BaS₂, then sulfur is 2 on right → so we need 2 Na₂S on left to get 2 S.

Then: BaF₂ + 2 Na₂S → BaS₂ + ? NaF

Left: Ba=1, F=2, Na=4, S=2
Right: Ba=1, S=2, so far good. Now Na and F: we have 4 Na and 2 F on left → so we need 4 NaF on right? But that would be 4 F, but we only have 2 F from BaF₂.

Problem: BaF₂ gives only 2 F, but if we make 4 NaF, we need 4 F.

This suggests maybe the product should be BaS, not BaS₂? But the worksheet says BaS₂.

Alternatively, perhaps it's a typo? But let’s assume it’s correct.

Wait — maybe we need 2 BaF₂?

Try: 2 BaF₂ + 2 Na₂S → 2 BaS₂ + 4 NaF? Let’s check:

Left: Ba=2, F=4, Na=4, S=2
Right: Ba=2, S=4 ← too many S! No.

Alternative approach: Maybe the intended product is BaS, not BaS₂? Because BaS₂ doesn't exist commonly.

Looking at common reactions: double displacement usually gives BaS and NaF.

Probably a typo in the worksheet — likely meant BaS.

Assume it’s BaS (not BaS₂). Then:

BaF₂ + Na₂S → BaS + 2 NaF

Left: Ba=1, F=2, Na=2, S=1
Right: Ba=1, S=1, Na=2, F=2 → Balanced!

Given that, and since BaS₂ is not standard, I think it’s safe to assume it’s a typo and should be BaS.

But the problem says BaS₂. Hmm.

Wait — let’s try to force it:

Suppose: BaF₂ + 2 Na₂S → BaS₂ + 4 NaF?
Left: Ba=1, F=2, Na=4, S=2
Right: Ba=1, S=2, Na=4, F=4 → F mismatch (left 2, right 4)

To fix F, need 2 BaF₂ → then F=4.

So: 2 BaF₂ + 2 Na₂S → 2 BaS₂ + 4 NaF?
Left: Ba=2, F=4, Na=4, S=2
Right: Ba=2, S=4 ← still wrong (S=4 vs 2)

No way to balance if product is BaS₂ unless we change coefficients differently.

Perhaps: BaF₂ + Na₂S → BaS + 2 NaF is intended.

I think there’s a typo. In most textbooks, this reaction is:

BaF₂ + Na₂S → BaS + 2 NaF

And that balances perfectly.

Since the worksheet says BaS₂, but that leads to imbalance, and given context, I’ll assume it’s a typo and proceed with BaS.

But to strictly follow the problem... let me see if any combination works.

Another idea: maybe it’s BaS₂ meaning barium disulfide, which might require different stoichiometry.

Suppose: 2 BaF₂ + 2 Na₂S → 2 BaS + 4 NaF — but again, that’s BaS.

I think for educational purpose, and since all other problems are straightforward, this is likely a typo.

I’ll go with the balanced version assuming BaS.

But the problem explicitly writes BaS₂. Let me search my knowledge — actually, barium polysulfides exist, like BaS₂, but they are less common.

In that case, perhaps:

Let’s set up equations.

Let a BaF₂ + b Na₂S → c BaS₂ + d NaF

Atoms:

Ba: a = c
F: 2a = d
Na: 2b = d
S: b = 2c

From Ba: a = c
From S: b = 2c = 2a
From Na: 2b = d → 2*(2a) = d → d = 4a
From F: 2a = d → 2a = 4a → 2a=4a → only if a=0, impossible.

Contradiction. So no solution if product is BaS₂.

Therefore, must be a typo. Intended product is BaS.

So I’ll solve as:

BaF₂ + Na₂S → BaS + 2 NaF

Which balances.

For the sake of completing the worksheet, I'll use that.

✔ Answer: 1, 1, 1, 2 *(assuming BaS instead of BaS₂)*

But since the problem says BaS₂, and it can't be balanced, perhaps the student should note that. However, for this exercise, I'll provide the answer based on common chemistry.

Alternatively, maybe it's BaS₂ and we need fractional coefficients? Not typical.

I think best to go with 1,1,1,2 assuming typo.

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10) NH₃ → N₂ + H₂

Left: N=1, H=3
Right: N=2, H=2

Need even numbers. Try 2 NH₃ → N₂ + 3 H₂

Left: N=2, H=6
Right: N=2, H=6 → Balanced!

✔ Answer: 2, 1, 3

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Now, compiling all answers:

Final Answer:

1) 1, 1, 1, 1
2) 1, 1, 1, 1
3) 2, 2, 1
4) 2, 1, 2, 1
5) 2, 1, 2, 1
6) 1, 2, 1, 1
7) 1, 2, 1, 1
8) 1, 6, 3, 2
9) 1, 1, 1, 2 *(assuming BaS instead of BaS₂ due to impossibility otherwise)*
10) 2, 1, 3

Note for #9: If the worksheet insists on BaS₂, it cannot be balanced with whole numbers. But in standard curriculum, it's likely a typo for BaS.

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Final Answer:
1) 1, 1, 1, 1
2) 1, 1, 1, 1
3) 2, 2, 1
4) 2, 1, 2, 1
5) 2, 1, 2, 1
6) 1, 2, 1, 1
7) 1, 2, 1, 1
8) 1, 6, 3, 2
9) 1, 1, 1, 2
10) 2, 1, 3