balancing chemical equations - practice 3
using your newly-found equation balancing knowledge, balance the following equations:
1) ___ br₂ + ___ lif → ___ libr + ___ f₂

2) ___ h₃po₄ + ___ fe(oh)₂ → ___ h₂o + ___ fe₃(po₄)₂

3) ___ c₃h₇oh + ___ o₂ → ___ co₂ + ___ h₂o

4) ___ ni(oh)₃ → ___ ni₂o₃ + ___ h₂o

5) ___ k₂so₃ + ___ mn(oh)₂ → ___ koh + ___ mnso₃

6) ___ naoh + ___ h₂so₄ → ___ h₂o + ___ na₂so₄

7) ___ li + ___ pb(oh)₂ → ___ pb + ___ lioh

Question

balancing chemical equations - practice 3
using your newly-found equation balancing knowledge, balance the following equations:
1) ___ br₂ + ___ lif → ___ libr + ___ f₂

2) ___ h₃po₄ + ___ fe(oh)₂ → ___ h₂o + ___ fe₃(po₄)₂

3) ___ c₃h₇oh + ___ o₂ → ___ co₂ + ___ h₂o

4) ___ ni(oh)₃ → ___ ni₂o₃ + ___ h₂o

5) ___ k₂so₃ + ___ mn(oh)₂ → ___ koh + ___ mnso₃

6) ___ naoh + ___ h₂so₄ → ___ h₂o + ___ na₂so₄

7) ___ li + ___ pb(oh)₂ → ___ pb + ___ lioh

Accepted Answer

### Problem 1:
# Explanation:
## Step1: Balance Br and F atoms.
On left: Br₂ (2 Br), LiF (1 F). On right: LiBr (1 Br), F₂ (2 F).
To balance Br: need 2 LiBr (2 Br). To balance F: need 2 LiF (2 F) (since F₂ has 2 F, LiF must have 2 F to balance). Then Br₂ is 1 (2 Br in 2 LiBr matches Br₂).
## Step2: Verify all atoms.
Br: 2 (Br₂) = 2 (LiBr). Li: 2 (LiF) = 2 (LiBr). F: 2 (LiF) = 2 (F₂).
# Answer:
1, 2, 2, 1

### Problem 2:
# Explanation:
## Step1: Balance PO₄³⁻ and Fe.
Right: Fe₃(PO₄)₂ has 2 PO₄³⁻ and 3 Fe. So left: H₃PO₄ needs 2 (2 PO₄³⁻), Fe(OH)₂ needs 3 (3 Fe).
## Step2: Balance H and O.
Left H: 2×3 (H₃PO₄) + 3×2 (Fe(OH)₂) = 6 + 6 = 12. Right H₂O: 12/2 = 6.
Left O: 2×4 (H₃PO₄) + 3×2 (Fe(OH)₂) = 8 + 6 = 14. Right O: 6 (H₂O) + 2×4 (Fe₃(PO₄)₂) = 6 + 8 = 14.
# Answer:
2, 3, 6, 1

### Problem 3:
# Explanation:
## Step1: Balance C and H.
C₃H₇OH has 3 C, 8 H. So CO₂ needs 3×2=6? Wait, C₃H₇OH: 3 C, 8 H (7 from C₃H₇ + 1 from OH), 2 O.
Let’s set C₃H₇OH coefficient as 2 (to make H even: 16 H). Then CO₂: 6 (3×2 C), H₂O: 8 (16 H / 2).
## Step2: Balance O.
Left O: 2×2 (C₃H₇OH) + O₂×x. Right O: 6×2 (CO₂) + 8×1 (H₂O) = 12 + 8 = 20. Left O: 4 + 2x = 20 → 2x=16 → x=9.
# Answer:
2, 9, 6, 8

### Problem 4:
# Explanation:
## Step1: Balance Ni and O/H.
Right: Ni₂O₃ has 2 Ni. So left Ni(OH)₃ needs 2 (2 Ni).
## Step2: Balance H and O.
Left H: 2×3 = 6 → H₂O needs 3 (6 H / 2). Left O: 2×3 = 6. Right O: 3 (Ni₂O₃) + 3 (H₂O) = 6.
# Answer:
2, 1, 3

### Problem 5:
# Explanation:
## Step1: Balance K and Mn.
K₂SO₃ has 2 K, Mn(OH)₂ has 1 Mn. Right KOH: 2 (to balance 2 K), MnSO₃: 1 (to balance Mn).
## Step2: Verify all atoms.
K: 2 (K₂SO₃) = 2 (KOH). Mn: 1 (Mn(OH)₂) = 1 (MnSO₃). SO₃²⁻: 1 (K₂SO₃) = 1 (MnSO₃). OH⁻: 2 (Mn(OH)₂ has 2 OH⁻? Wait, Mn(OH)₂ has 2 OH⁻, KOH has 2 OH⁻ (2 KOH). H: 2 (Mn(OH)₂) = 2 (KOH). O: 3 (K₂SO₃) + 2 (Mn(OH)₂) = 3 + 2 = 5; right: 2 (KOH) + 3 (MnSO₃) = 2 + 3 = 5.
# Answer:
1, 1, 2, 1

### Problem 6:
# Explanation:
## Step1: Balance Na and SO₄²⁻.
Na₂SO₄ has 2 Na, so NaOH needs 2. H₂SO₄ has 1 SO₄²⁻, so Na₂SO₄ is 1.
## Step2: Balance H and O.
Left H: 2×1 (NaOH) + 2×1 (H₂SO₄) = 2 + 2 = 4. Right H₂O: 4/2 = 2.
Left O: 2×1 (NaOH) + 4×1 (H₂SO₄) = 2 + 4 = 6. Right O: 2×1 (H₂O) + 4×1 (Na₂SO₄) = 2 + 4 = 6.
# Answer:
2, 1, 2, 1

### Problem 7:
# Explanation:
## Step1: Balance OH⁻ and Li.
Pb(OH)₂ has 2 OH⁻, so LiOH needs 2 (2 OH⁻). Then Li needs 2 (2 LiOH).
## Step2: Balance Pb.
Left Pb(OH)₂: 1, right Pb: 1.
# Answer:
2, 1, 1, 2

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QUESTION IMAGE

Question

Explanation:

Problem 1:

Step1: Balance Br and F atoms.

Step2: Verify all atoms.

Step1: Balance PO₄³⁻ and Fe.

Step2: Balance H and O.

Step1: Balance C and H.

Step2: Balance O.

Answer:

Problem 2: