The volume of chlorine gas at $S T P$ (in $L i t r e s$ ) required to completely react with 5 moles of hot and concentrated $N a O H$ solution is:

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Solution

When chlorine gas reacts with hot and concentrated $N a O H$ solution, it disproportionates into $C l^{-}$ and $C l O_{3}^{-}$ ions.

$3 C l_{2} + 6 N a O H \to 5 N a C l + N a C l O_{3} + 3 H_{2} O$

The moles of $C l_{2}$ gas required to react with 5 moles of $N a O H$ = 2.5 mol
So, the volume of $C l_{2}$ gas required at $S T P$ is 56 $L$

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The volume of chlorine gas at STP (in Litres) required to completely react with 5 moles of hot and concentrated NaOH solution is:

When chlorine gas reacts with hot and concentrated NaOH solution, it disproportionates into Cl− and ClO−3 ions. 3Cl2+6NaOH→5NaCl+NaClO3+3H2O The moles of Cl2 gas required to react with 5 moles of NaOH = 2.5 mol So, the volume of Cl2 gas required at STP is 56 L

The volume of chlorine gas at $S T P$ (in $L i t r e s$ ) required to completely react with 5 moles of hot and concentrated $N a O H$ solution is: