1. **State the problem:** We need to find the limiting reactant in the reaction: $$\text{CCl}_4 + 2\text{HF} \rightarrow \text{CCl}_2\text{F}_2 + 2\text{HCl}$$ Given moles: - CCl\(_4\): 0.0195 mol - HF: 0.150 mol 2. **Write the mole ratio from the balanced equation:** $$1 \text{ mol CCl}_4 : 2 \text{ mol HF}$$ 3. **Calculate the required moles of HF to react with given CCl\(_4\):** $$0.0195 \text{ mol CCl}_4 \times \frac{2 \text{ mol HF}}{1 \text{ mol CCl}_4} = 0.0390 \text{ mol HF}$$ 4. **Compare the available moles of HF with required moles:** - Available HF = 0.150 mol - Required HF = 0.0390 mol Since available HF (0.150 mol) > required HF (0.0390 mol), HF is in excess. 5. **Determine the limiting reactant:** - CCl\(_4\) requires 0.0390 mol HF but we have more HF than needed. - Therefore, CCl\(_4\) is the limiting reactant because it will be completely consumed first. **Final answer:** The limiting reactant is **CCl\(_4\)**.