What is the final volume of reaction mixture if 8 L of H2 and 6L of Cl2 are allowed to react.

Question

Answer 1

The correct answer is (3) 14 L

The equation for the reaction is:

H₂(g) + Cl₂(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 H₂ (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 H₂ requires = 22.7 L of Cl₂
So, 8 L of H₂ will require = 8 L of Cl₂
But we have only 6 L of Cl₂. Thus, Chlorine gas is the limiting reagent and the amount of products will depend on its amount only.

Now, 22.7 of Cl₂ forms = 45.4 L of HCl,
So, 6 L of Cl₂ 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 H₂.
So, total volume of the reaction mixture will be 14 L.