Volume of Cl₂ = 9.34 L
Further explanation
Reaction
2NaI + Cl₂ ⇒ 2NaCl + I₂
mol NaI (MW=149,89 g/mol) :
[tex]\tt \dfrac{125}{149,89}=0.834[/tex]
mol Cl₂ : mol NaI = 1 : 2
mol Cl₂ :
[tex]\tt \dfrac{1}{2}\times 0.834=0.417[/tex]
Assumptions at STP(1 mol=22.4 L) :
[tex]\tt 0.417\times 22.4~L=9.34~L[/tex]
