One mole of a mixture of $C O$ and $C O_{2}$ requires exactly $20 g$ of $N a O H$ to convert all the $C O_{2}$ into $N a_{2} C O_{3}$ . How many more grams of $N a O H$ would it require for conversion into $N a_{2} C O_{3}$ if the mixture (one mole) is completely oxidised to $C O_{2}$ ?

60 g

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80 g

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40 g

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20 g

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Solution

The correct option is A $60 g$
Let moles of $C O$ $= x$
So, moles of $C O_{2}$ $= 1 - x$

Now,
$2 N a O H + C O_{2} \to N a_{2} C O_{3} + H_{2} O$

2 moles of $N a O H$ will react with 1 mole of $C O_{2}$

So, 20g $N a O H$ = 0.5moles $N a O H$ will react with $\frac{1}{4} = 0.25 m o l e s$ of $C O_{2}$

Therefore, $x = 0.75 m o l e s$ of $C O$
Now,
On oxidation $C O + \frac{1}{2} \times O_{2} \to C O_{2}$

$C O (0.75 m o l e s) \to C O_{2} (0.75 m o l e s)$

Total moles of of $C O_{2}$ = (0.25 + 0.75)moles = 1 moles

Therefore, 1 mole of $C O_{2}$ will react with 2 moles of $N a O H$ i.e it will require $(40 \times 2) = 80 g$ of $N a O H$

So, 60g more is required.

Hence, the correct option is $A$

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