Free Printable Balancing and Classifying Chemical Equations Worksheets - Free Printable

Explanation:
We are balancing double displacement reactions. In these reactions, the positive ions (cations) and negative ions (anions) swap partners. The key is to make sure the number of atoms of each element is the same on both sides — and also that the total charge is balanced (though for neutral compounds, we just ensure atom counts match).

Let’s go one by one carefully.

(a) NaBr + H₃PO₄ → Na₃PO₄ + HBr
- On right: Na₃PO₄ has 3 Na⁺, PO₄³⁻; HBr has H⁺ and Br⁻.
- Left: NaBr gives Na⁺ and Br⁻; H₃PO₄ gives 3H⁺ and PO₄³⁻.
To get 3 Na on right, need 3 NaBr on left. That gives 3 Br⁻, so need 3 HBr on right.
Check H: left has 3H from H₃PO₄; right has 3H from 3 HBr → OK.
✔ Balanced: 3 NaBr + 1 H₃PO₄ → 1 Na₃PO₄ + 3 HBr

(b) Ca(OH)₂ + Al₂(SO₄)₃ → CaSO₄ + Al(OH)₃
- Ca(OH)₂: Ca²⁺, 2 OH⁻
- Al₂(SO₄)₃: 2 Al³⁺, 3 SO₄²⁻
Products: CaSO₄ (Ca²⁺ + SO₄²⁻), Al(OH)₃ (Al³⁺ + 3 OH⁻)
Need 3 Ca²⁺ to pair with 3 SO₄²⁻ → 3 CaSO₄
That needs 3 Ca(OH)₂ → gives 6 OH⁻
Each Al(OH)₃ needs 3 OH⁻, so 6 OH⁻ makes 2 Al(OH)₃
And Al₂(SO₄)₃ already has 2 Al³⁺ → matches 2 Al(OH)₃
So: 3 Ca(OH)₂ + 1 Al₂(SO₄)₃ → 3 CaSO₄ + 2 Al(OH)₃

(c) AgI + Fe₂(CO₃)₃ → FeI₃ + Ag₂CO₃
Fe₂(CO₃)₃: 2 Fe³⁺, 3 CO₃²⁻
AgI: Ag⁺, I⁻
Products: FeI₃ (Fe³⁺ + 3 I⁻), Ag₂CO₃ (2 Ag⁺ + CO₃²⁻)
We need 3 CO₃²⁻ → 3 Ag₂CO₃ → needs 6 Ag⁺ → so 6 AgI
6 AgI gives 6 I⁻ → FeI₃ needs 3 I⁻ per Fe, so 6 I⁻ → 2 FeI₃
Fe₂(CO₃)₃ gives 2 Fe³⁺ → matches 2 FeI₃
✔ So: 6 AgI + 1 Fe₂(CO₃)₃ → 2 FeI₃ + 3 Ag₂CO₃

(d) CoBr₃ + CaSO₄ → CaBr₂ + Co₂(SO₄)₃
CoBr₃: Co³⁺, 3 Br⁻
CaSO₄: Ca²⁺, SO₄²⁻
Right: CaBr₂ (Ca²⁺ + 2 Br⁻), Co₂(SO₄)₃ (2 Co³⁺, 3 SO₄²⁻)
To get Co₂(SO₄)₃, need 2 Co³⁺ and 3 SO₄²⁻
So need 2 CoBr₃ (gives 2 Co³⁺, 6 Br⁻)
Need 3 CaSO₄ (gives 3 Ca²⁺, 3 SO₄²⁻)
Now right side: Co₂(SO₄)₃ uses 2 Co, 3 SO₄
Remaining: 3 Ca²⁺ and 6 Br⁻ → forms 3 CaBr₂ (each uses 1 Ca²⁺ and 2 Br⁻)
Yes! ✔
2 CoBr₃ + 3 CaSO₄ → 3 CaBr₂ + 1 Co₂(SO₄)₃

(e) Na₃P + CaF₂ → NaF + Ca₃P₂
Na₃P: 3 Na⁺, P³⁻
CaF₂: Ca²⁺, 2 F⁻
Products: NaF (Na⁺ + F⁻), Ca₃P₂ (3 Ca²⁺, 2 P³⁻)
To get Ca₃P₂, need 2 P³⁻ → need 2 Na₃P → gives 6 Na⁺, 2 P³⁻
Ca₃P₂ needs 3 Ca²⁺ → need 3 CaF₂ → gives 3 Ca²⁺, 6 F⁻
6 Na⁺ + 6 F⁻ → 6 NaF
✔ So: 2 Na₃P + 3 CaF₂ → 6 NaF + 1 Ca₃P₂

(f) Li₃PO₄ + NaBr → Na₃PO₄ + LiBr
Li₃PO₄: 3 Li⁺, PO₄³⁻
NaBr: Na⁺, Br⁻
Right: Na₃PO₄ (3 Na⁺, PO₄³⁻), LiBr (Li⁺, Br⁻)
Swap: need 3 Na⁺ → 3 NaBr
That gives 3 Br⁻ → need 3 LiBr
Left: Li₃PO₄ gives 3 Li⁺ → matches 3 LiBr
✔ 1 Li₃PO₄ + 3 NaBr → 1 Na₃PO₄ + 3 LiBr

(g) H₂SO₄ + NaNO₂ → HNO₂ + Na₂SO₄
H₂SO₄: 2 H⁺, SO₄²⁻
NaNO₂: Na⁺, NO₂⁻
Right: HNO₂ (H⁺, NO₂⁻), Na₂SO₄ (2 Na⁺, SO₄²⁻)
To get Na₂SO₄, need 2 Na⁺ → 2 NaNO₂
2 NaNO₂ gives 2 NO₂⁻ → need 2 HNO₂
H₂SO₄ gives 2 H⁺ + SO₄²⁻ → matches 2 HNO₂ + Na₂SO₄
✔ 1 H₂SO₄ + 2 NaNO₂ → 2 HNO₂ + 1 Na₂SO₄

(h) NaBr + CaF₂ → NaF + CaBr₂
NaBr: Na⁺, Br⁻
CaF₂: Ca²⁺, 2 F⁻
Products: NaF (Na⁺, F⁻), CaBr₂ (Ca²⁺, 2 Br⁻)
Need 2 Na⁺ and 2 Br⁻ for CaBr₂ → 2 NaBr
2 NaBr gives 2 Na⁺, 2 Br⁻
CaF₂ gives Ca²⁺, 2 F⁻ → need 2 NaF
So: 2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂

(i) AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃
AlBr₃: Al³⁺, 3 Br⁻
K₂SO₄: 2 K⁺, SO₄²⁻
Right: KBr (K⁺, Br⁻), Al₂(SO₄)₃ (2 Al³⁺, 3 SO₄²⁻)
Need 2 Al³⁺ → 2 AlBr₃ → gives 6 Br⁻
Need 3 SO₄²⁻ → 3 K₂SO₄ → gives 6 K⁺
6 K⁺ + 6 Br⁻ → 6 KBr
✔ 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃

(j) FeCl₃ + NaOH → Fe(OH)₃ + NaCl
FeCl₃: Fe³⁺, 3 Cl⁻
NaOH: Na⁺, OH⁻
Fe(OH)₃: Fe³⁺, 3 OH⁻
So need 3 OH⁻ → 3 NaOH
3 NaOH gives 3 Na⁺, 3 OH⁻
3 Na⁺ + 3 Cl⁻ → 3 NaCl
✔ 1 FeCl₃ + 3 NaOH → 1 Fe(OH)₃ + 3 NaCl

(k) CoBr₃ + CaSO₄ → CaBr₂ + Co₂(SO₄)₃
This is same as (d), just different metal. Same logic:
CoBr₃ → Co³⁺, 3 Br⁻
CaSO₄ → Ca²⁺, SO₄²⁻
Product Co₂(SO₄)₃ needs 2 Co³⁺, 3 SO₄²⁻ → 2 CoBr₃, 3 CaSO₄
Then 3 Ca²⁺ + 6 Br⁻ → 3 CaBr₂
✔ 2 CoBr₃ + 3 CaSO₄ → 3 CaBr₂ + 1 Co₂(SO₄)₃

(l) Li₃PO₄ + NaBr → Na₃PO₄ + LiBr
Same as (f):
Li₃PO₄ + 3 NaBr → Na₃PO₄ + 3 LiBr
✔ 1 Li₃PO₄ + 3 NaBr → 1 Na₃PO₄ + 3 LiBr

(m) CaF₂ + Li₂SO₄ → CaSO₄ + LiF
CaF₂: Ca²⁺, 2 F⁻
Li₂SO₄: 2 Li⁺, SO₄²⁻
Products: CaSO₄ (Ca²⁺, SO₄²⁻), LiF (Li⁺, F⁻)
Need 2 F⁻ → 2 LiF → need 2 Li⁺ → Li₂SO₄ provides 2 Li⁺
So: 1 CaF₂ + 1 Li₂SO₄ → 1 CaSO₄ + 2 LiF
✔ 1 CaF₂ + 1 Li₂SO₄ → 1 CaSO₄ + 2 LiF

(n) RbNO₃ + BeF₂ → Be(NO₃)₂ + RbF
RbNO₃: Rb⁺, NO₃⁻
BeF₂: Be²⁺, 2 F⁻
Be(NO₃)₂: Be²⁺, 2 NO₃⁻
So need 2 NO₃⁻ → 2 RbNO₃
2 RbNO₃ gives 2 Rb⁺, 2 NO₃⁻
2 Rb⁺ + 2 F⁻ → 2 RbF
✔ 2 RbNO₃ + 1 BeF₂ → 1 Be(NO₃)₂ + 2 RbF

(o) MgF₂ + Li₂CO₃ → MgCO₃ + LiF
MgF₂: Mg²⁺, 2 F⁻
Li₂CO₃: 2 Li⁺, CO₃²⁻
Products: MgCO₃ (Mg²⁺, CO₃²⁻), LiF (Li⁺, F⁻)
Need 2 F⁻ → 2 LiF → need 2 Li⁺ → Li₂CO₃ gives exactly that
So: 1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF
✔ 1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF

Now let’s compile all final coefficients:

(a) 3, 1, 1, 3
(b) 3, 1, 3, 2
(c) 6, 1, 2, 3
(d) 2, 3, 3, 1
(e) 2, 3, 6, 1
(f) 1, 3, 1, 3
(g) 1, 2, 2, 1
(h) 2, 1, 2, 1
(i) 2, 3, 6, 1
(j) 1, 3, 1, 3
(k) 2, 3, 3, 1
(l) 1, 3, 1, 3
(m) 1, 1, 1, 2
(n) 2, 1, 1, 2
(o) 1, 1, 1, 2

Double-check a few tricky ones:

- (e): 2 Na₃P = 6 Na, 2 P; 3 CaF₂ = 3 Ca, 6 F → products: 6 NaF (6 Na, 6 F), Ca₃P₂ (3 Ca, 2 P) ✔
- (i): 2 AlBr₃ = 2 Al, 6 Br; 3 K₂SO₄ = 6 K, 3 SO₄; products: 6 KBr (6 K, 6 Br), Al₂(SO₄)₃ (2 Al, 3 SO₄) ✔
- (n): 2 RbNO₃ = 2 Rb, 2 NO₃; BeF₂ = Be, 2 F; products: Be(NO₃)₂ (Be, 2 NO₃), 2 RbF (2 Rb, 2 F) ✔

All correct.

Final Answer:
(a) 3, 1, 1, 3
(b) 3, 1, 3, 2
(c) 6, 1, 2, 3
(d) 2, 3, 3, 1
(e) 2, 3, 6, 1
(f) 1, 3, 1, 3
(g) 1, 2, 2, 1
(h) 2, 1, 2, 1
(i) 2, 3, 6, 1
(j) 1, 3, 1, 3
(k) 2, 3, 3, 1
(l) 1, 3, 1, 3
(m) 1, 1, 1, 2
(n) 2, 1, 1, 2
(o) 1, 1, 1, 2