Step 1: Write the balanced equation and note which product is not a gas.
\[ CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(l) + 4HCl(g) \]
$CCl_4$ is a liquid at STP, so whatever amount of it forms drops out of the gas volume completely. Only unreacted $CH_4$, unreacted $Cl_2$, and the $HCl$ formed count towards the final gas volume.
Step 2: Let $x$ be the volume of $CH_4$ that reacts.
We are told $50\%$ of the $1$ L of $CH_4$ reacts, so:
\[ x = 0.5 \, \text{L} \]
Step 3: Use the mole ratio to track everything else in terms of $x$.
From the equation, $1$ volume of $CH_4$ needs $4$ volumes of $Cl_2$ and produces $4$ volumes of $HCl$:
\[ Cl_2 \, \text{used} = 4x, \quad HCl \, \text{formed} = 4x \]
Step 4: Check that the given $Cl_2$ amount matches this exactly.
\[ 4x = 4(0.5) = 2 \, \text{L} \]
This is precisely the $2$ L of $Cl_2$ given, so all the chlorine is used up with nothing left over, and the stoichiometry checks out.
Step 5: Add up what remains as gas.
Unreacted $CH_4$ left is $1-x = 0.5$ L, unreacted $Cl_2$ left is $0$ L, and $HCl$ formed is $4x=2$ L. The liquid $CCl_4$ is excluded.
\[ \text{Total gas volume} = (1-x) + 0 + 4x = 1 + 3x \]
Step 6: Substitute $x=0.5$.
\[ \text{Total gas volume} = 1 + 3(0.5) = 1 + 1.5 = 2.5 \, \text{L} \]
Final Answer:
\[ \boxed{2.5 \, \text{L}} \]