6. $$ \_ $$Ca $$ +\ldots \mathrm{HoF}_{3} \rightarrow \ldots \mathrm{Ho}+\ldots \mathrm{CaF}_{2} $$

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To balance the chemical reaction involving calcium (Ca) and holmium trifluoride (HoF₃), we need to identify the products and ensure that the number of atoms of each element is the same on both sides of the equation.

### Step 1: Identify the Reactants and Products
- Reactants: $$ \text{Ca} $$ and $$ \text{HoF}_3 $$
- Products: $$ \text{Ho} $$ and $$ \text{CaF}_2 $$

### Step 2: Write the Unbalanced Equation
The unbalanced equation can be written as:
$$
\text{Ca} + \text{HoF}_3 \rightarrow \text{Ho} + \text{CaF}_2
$$

### Step 3: Balance the Equation
1. **Calcium (Ca)**: There is 1 Ca on both sides.
2. **Holmium (Ho)**: There is 1 Ho on the product side, so we need 1 HoF₃ on the reactant side.
3. **Fluorine (F)**: There are 3 F in HoF₃, and we need to balance this with CaF₂, which has 2 F. To balance the fluorine, we can use 3 HoF₃ and 1.5 CaF₂, but since we cannot have half a molecule, we will multiply the entire equation by 2.

### Step 4: Write the Balanced Equation
After balancing, the equation becomes:
$$
2 \text{Ca} + 6 \text{HoF}_3 \rightarrow 2 \text{Ho} + 3 \text{CaF}_2
$$

### Final Balanced Equation
The final balanced equation is:
$$
2 \text{Ca} + 6 \text{HoF}_3 \rightarrow 2 \text{Ho} + 3 \text{CaF}_2
$$

This ensures that the number of each type of atom is the same on both sides of the equation.
The balanced chemical equation is: $$ 2 \text{Ca} + 6 \text{HoF}_3 \rightarrow 2 \text{Ho} + 3 \text{CaF}_2 $$

6. \( \_ \)Ca \( +\ldots \mathrm{HoF}_{3} \rightarrow \ldots \mathrm{Ho}+\ldots \mathrm{CaF}_{2} \)

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