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Oxidation of Alcohols

Consider the ...

Question

Consider the following series of reactions:
$C l_{2} + 2 N a O H ⟶ N a C l + N a C l O + H_{2} O 3 N a C l O ⟶ 2 N a C l + N a C l O_{3} 4 N a C l O_{3} ⟶ N a C l + 3 N a C l O_{4}$

How many mole of total $N a C l$ formed by using $1$ mole $C l_{2}$ and other reagents in excess?

\frac{1}{12} mole

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1.67 mole

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1.75 mole

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0.75 mole

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Solution

The correct option is C $1.75 mole$
Given reactions are:
$C l_{2} + 2 N a O H ⟶ N a C l + N a C l O + H_{2} O 3 N a C l O ⟶ 2 N a C l + N a C l O_{3} 4 N a C l O_{3} ⟶ N a C l + 3 N a C l O_{4}$
$1$ mole of $C l_{2}$ reacts with $2$ moles of $N a O H$ to produce $1$ mole of $N a C l$ and $1$ mole of $N a C l O$ .
Again, $3$ moles of $N a C l O$ gives $2$ moles $N a C l$ and $1$ mole $N a C l O_{3}$ .
So $1$ mole of $N a C l O$ gives $\frac{2}{3}$ moles $N a C l$ and $\frac{1}{3}$ mole $N a C l O_{3}$ .
Again $4$ moles of $N a C l O_{3}$ gives $1$ mole $N a C l$ .
So $\frac{1}{3}$ moles of $N a C l O_{3}$ gives $\frac{1}{12}$ mole $N a C l$ .
So total $N a C l$ produced
$= 1 + \frac{2}{3} + \frac{1}{12} = \frac{12 + 8 + 1}{12} = \frac{21}{3} = 1.75$

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Similar questions

C l_{2} + 2 N a O H ⟶ N a C l + N a C l O + H_{2} O 3 N a C l O ⟶ 2 N a C l + N a C l O_{3} 4 N a C l O_{3} ⟶ N a C l + 3 N a C l O_{4}

N a C l O_{3}

C l_{2}

C l_{2} + 2 N a O H \to N a C l + N a C l O + H_{2} O

3 N a C l O \to 2 N a C l + N a C l O_{3}

4 N a C l O_{3} \to 3 N a C l O_{4} + N a C l

C l_{2}

106.5

N a C l O_{3}

I) C l_{2} + 2 N a O H \to N a C l + N a C l O + H_{2} O

I I) 3 N a C l O \to 2 N a C l + N a C l O_{3}

I I I) 4 N a C l O_{3} \to 3 N a C l O_{4} + N a C l

C l_{2}

N a C l O_{4}

I) C l_{2} + 2 N a O H \to N a C l + N a C l O + H_{2} O

I I) 3 N a C l O \to 2 N a C l + N a C l O_{3}

I I I) 4 N a C l O_{3} \to 3 N a C l O_{4} + N a C l

C l_{2}

N a C l O_{4}

2 N a O H + C l_{2} \to N a C l +

+ H_{2} O

N a_{2} S O_{3} + 2 H C l \to 2 N a C l + H_{2} O +

C l_{2} + 3 F_{2} 573 K - -- \to

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