1 Answer

Dave Draper · Answer 1 · 2021-10-01T23:16:09+0000

Volume of Cl₂ = 9.34 L

Reaction

2NaI + Cl₂ ⇒ 2NaCl + I₂

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

$\tt (125)/(149,89)=0.834$

mol Cl₂ : mol NaI = 1 : 2

mol Cl₂ :

$\tt (1)/(2)* 0.834=0.417$

Assumptions at STP(1 mol=22.4 L) :

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

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