0.8 mole of a mixture of $C O$ and $C O_{2}$ requires exactly 40 gram of $N a O H$ in solution for complete conversion of all the $C O_{2}$ into $N a_{2} C O_{3}$ . How many moles more of $N a O H$ would it require for conversion into $N a_{2} C O_{3}$ , if the mixture (0.8 mole) is completely oxidised to $C O_{2}$ :

0.2

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0.6

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1.5

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Solution

The correct option is B 0.6
$2 N a O H + C O_{2} \to N a_{2} C O_{3} + H_{2} O$

Molar ratio $2$ $:$ $1$

$40 g N a O H = 1 m o l e N a O H = \frac{1}{2} m o l e C O_{2}$

(According to molar ratio)

Total mixture of $(C O + C O_{2}) = 0.8 m o l e$

$\to 2 C O + O_{2} \to 2 C O_{2}$ shows us that the ratio of $C O : C O_{2}$ is $1 : 1$ . Therefore, if mixture was completely oxidized then there would be $0.8 m o l e C O_{2}$ in total.

$\to$ Of $0.8 m o l e$ we have already used $0.5 m o l e$

$\to$ We have $08 - 0.5 = 0.3 m o l e C O_{2}$

$1 m o l e C O_{2} \to 2 m o l e s N a O H$

$∴ 0.3 m o l e s C O_{2} \to 0.6 m o l e s N a O H$

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