Question

What volume in liters of chlorine gas would be fully required to react to 125 g of Nal

Answer 1

Volume of Cl₂ = 9.34 L

Further explanation

Reaction

2NaI + Cl₂ ⇒ 2NaCl + I₂

mol NaI (MW=149,89 g/mol) :

$\tt \dfrac{125}{149,89}=0.834$

mol Cl₂ : mol NaI = 1 : 2

mol Cl₂ :

$\tt \dfrac{1}{2}\times 0.834=0.417$

Assumptions at STP(1 mol=22.4 L) :

$\tt 0.417\times 22.4~L=9.34~L$