Step-by-step guide to balancing chemical equations.

Let's solve the balancing chemical equations problem step by step, focusing on the main reaction:

Unbalanced Equation:

$$
\text{Na}_3\text{PO}_4 + \text{CaCl}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{NaCl}
$$

We need to balance this equation so that the number of atoms of each element is the same on both sides.

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🔍 Step 1: Identify all elements involved

- Sodium (Na)
- Phosphorus (P)
- Oxygen (O)
- Calcium (Ca)
- Chlorine (Cl)

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🔍 Step 2: Count atoms on each side (unbalanced)

| Element | Left Side | Right Side |
|--------|----------|-----------|
| Na | 3 | 1 |
| P | 1 | 2 |
| O | 4 | 8 |
| Ca | 1 | 3 |
| Cl | 2 | 1 |

We see imbalances in all elements.

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🔍 Step 3: Start balancing with the most complex compound

The most complex compound is Ca₃(PO₄)₂, which contains:
- 3 Ca
- 2 PO₄ groups → so 2 P and 8 O

So we want 2 P on the left, but currently only 1 P per Na₃PO₄.

👉 So we put a 2 in front of Na₃PO₄:

$$
2\text{Na}_3\text{PO}_4 + \text{CaCl}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{NaCl}
$$

Now check atoms again:

| Element | Left Side | Right Side |
|--------|----------|-----------|
| Na | 2×3 = 6 | 1 |
| P | 2 | 2 ✔ |
| O | 2×4 = 8 | 8 ✔ |
| Ca | 1 | 3 |
| Cl | 2 | 1 |

Now P and O are balanced. But we still have issues with Ca and Cl.

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🔍 Step 4: Balance Ca

We need 3 Ca on the right, but only 1 on the left.

So place a 3 in front of CaCl₂:

$$
2\text{Na}_3\text{PO}_4 + 3\text{CaCl}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{NaCl}
$$

Now:
- Ca: 3 on both sides ✔
- Cl: 3×2 = 6 on left, but only 1 on right → need 6 Cl on right

So put a 6 in front of NaCl:

$$
2\text{Na}_3\text{PO}_4 + 3\text{CaCl}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\text{NaCl}
$$

Now check Na:
- Left: 2×3 = 6 Na
- Right: 6×1 = 6 Na ✔

Check Cl:
- Left: 3×2 = 6 Cl
- Right: 6×1 = 6 Cl ✔

Now double-check all:

| Element | Left Side | Right Side |
|--------|----------|-----------|
| Na | 6 | 6 ✔ |
| P | 2 | 2 ✔ |
| O | 8 | 8 ✔ |
| Ca | 3 | 3 ✔ |
| Cl | 6 | 6 ✔ |

✔ All atoms are balanced!

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✔ Final Balanced Equation:

$$
\boxed{2\text{Na}_3\text{PO}_4 + 3\text{CaCl}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\text{NaCl}}
$$

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📌 Bonus: The second equation shown in blue

$$
\text{S}_8 + 24\text{F}_2 \rightarrow 8\text{SF}_6
$$

Let’s verify if this is balanced.

- Left: S = 8, F = 24×2 = 48
- Right: S = 8, F = 8×6 = 48

✔ Yes, it is balanced!

This is a correct balancing for sulfur reacting with fluorine to form sulfur hexafluoride.

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✔ Summary:

- Main equation:
$$
2\text{Na}_3\text{PO}_4 + 3\text{CaCl}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\text{NaCl}
$$

- Second equation (already balanced):
$$
\text{S}_8 + 24\text{F}_2 \rightarrow 8\text{SF}_6
$$

Let me know if you'd like a visual explanation or tips for balancing other types of reactions!