**The correct answer is (3) 14 L**

The equation for the reaction is:

H_{2}(g) + Cl_{2}(g) → 2HCl

According to the mole concept, under standard conditions, one mole of any gas occupies 22.7 L volume.

So, the equation says that 1 mole of H_{2} (22.7 L) requires 1 mole of Cl_{2} (22.7 L) to form 2 moles of HCl (45.4 L)

Thus, 22.7 L of H_{2} requires = 22.7 L of Cl_{2}

So, 8 L of H_{2} will require = 8 L of Cl_{2}

But we have only 6 L of Cl_{2}. Thus, Chlorine gas is the limiting reagent and the amount of products will depend on its amount only.

Now, 22.7 of Cl_{2} forms = 45.4 L of HCl,

So, 6 L of Cl_{2} will form = 45.4/22.7×6 = 12 L of HCl

Now, chlorine gas is completely consumed as it is the limiting reagent. Along with it, 6L of hydrogen gas is also consumed. Now, 2L of hydrogen gas is still present.

Thus, the reaction mixture will contain 12 L of HCl and 2 L of H_{2}.

**So, total volume of the reaction mixture will be 14 L.**