Complete the following reaction:
`SnCl_(4)+C_(2)H_(5)Cl+Na to`

Answer

To complete the reaction between SnCl₄, C₂H₅Cl, and Na, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in the reaction are:
- Tin(IV) chloride (SnCl₄)
- Ethyl chloride (C₂H₅Cl)
- Sodium (Na)

### Step 2: Determine the Reaction Type
This reaction is a combination reaction where sodium reacts with tin(IV) chloride and ethyl chloride. Sodium will act as a reducing agent.

### Step 3: Write the Chemical Reaction
The reaction can be written as:
\[ \text{SnCl}_4 + \text{C}_2\text{H}_5\text{Cl} + \text{Na} \rightarrow \text{Na}_2\text{SnCl}_6 + \text{C}_4\text{H}_{10} \]

### Step 4: Balance the Reaction
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides.

1. **Tin (Sn)**: 1 on both sides.
2. **Sodium (Na)**: We have 2 Na on the product side, so we need 2 Na on the reactant side.
3. **Chlorine (Cl)**: We have 6 Cl in Na₂SnCl₆ and 1 Cl in C₂H₅Cl, totaling 7 Cl on the product side. Therefore, we need 6 Cl from SnCl₄ and 1 Cl from C₂H₅Cl, which is correct.
4. **Carbon (C)**: We have 2 C in C₂H₅Cl and 2 C in C₄H₁₀ (as it contains 4 C).
5. **Hydrogen (H)**: We have 5 H in C₂H₅Cl and 10 H in C₄H₁₀ (as it contains 10 H).

The balanced equation becomes:
\[ 2 \text{C}_2\text{H}_5\text{Cl} + \text{SnCl}_4 + 2 \text{Na} \rightarrow \text{Na}_2\text{SnCl}_6 + \text{C}_4\text{H}_{10} \]

### Final Balanced Reaction
Thus, the complete and balanced reaction is:
\[ 2 \text{C}_2\text{H}_5\text{Cl} + \text{SnCl}_4 + 2 \text{Na} \rightarrow \text{Na}_2\text{SnCl}_6 + \text{C}_4\text{H}_{10} \]

---