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

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.